Answer:
D
Step-by-step explanation:
the equation of a line in point- slope form is
y - b = m(x - a)
where m is the slope and (a, b ) a point on the line
here m = 2 and (a, b ) = (- 2, 1 ) , then
y - 1 = 2(x - (- 2)) , that is
y - 1 = 2(x + 2)
Is this congruent yes or no explain why
Answer:
Yes, they are congruent. They must be congruent because you know that 3 out of the 3 angles are identical between the two triangles. You know this because both triangles have a right angle (90°) and another identical angle (the two curves). This means that the 3rd angle for both triangles must be identical as well. Therefore, the triangles are the same.
Cosmo’s family has a pool like this.
A pool is a cylinder, height 120 centimetre and width 5.4 metre.
What is the volume of the pool?
How many litres of water will the pool hold?
How long will it take to fill the pool at a rate of 50 L/min?
a. The volume of the pool is 27.483 m³
b. The number of litres of water the pool will hold is 27483 L
c. The pool will fill at a rate of 50 L/min in 549.7 min
The pool is a cylinder, so, we need to find its volume
a. What is volume of the pool?Since the pool is a cylinder, its volume is, V = πd²h/4 where
d = width of pool = 5.4 m and h = height of pool = 120 cm = 1.20 mSo, V = πd²h/4
= π(5.4 m)² × 1.20 m/4
= π × 29.16 m² × 1.20 m/4
= 34.992π m³/4
= 8.748π m³
= 27.483 m³
So, the volume of the pool is 27.483 m³
b. How many litres of water will the pool hold?Since the volume of the pool, V = 27.483 m³ and 1000 Litres = 1 m³
So, the volume of the pool in litres is V = 27.483 m³
= 27.483 × 1 m³
= 27.483 × 1 000 L
= 27483 L
So, the number of litres of water the pool will hold is 27483 L
c. How long will it take to fill the pool at a rate of 50 L/min?Since the volume of the pool is V = 27483 L, the time it will take the pool to fill at a rate of 50 L/min is given by t = Volume/rate
= 27483 L/50 L/min
= 549.66 min
≅ 549.7 min
So, how long will it take to fill the pool at a rate of 50 L/min is 549.7 min
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There are 60 feet between the bases on a softball diamond. How far does a catcher throw drone home plate to second base
Answer: [tex]12\sqrt{5}[/tex]
Required theorem:
Pythagorean theorem = [tex]a^2 + b^2=c^2[/tex]
In other words, one side length squared + another side length squared = the middle lined squared.
Step-by-step explanation:
[tex]60^2+60^2 = c^2\\\\360+360=c^2\\\\720=c^2\\\\\ \sqrt{c^2} =\sqrt{720}\\\\c=+/- 12\sqrt{5}[/tex]
We can only have a positive distance so your final answer is :
[tex]12\sqrt{5}[/tex]
f(x) = x². What is g(x)? g(x) -5
A. g(x)=x²
B. g(x) = -2x²
C. g(x) = x² - 2
D. g(x) = -x² - 2
Answer:
D. g(x) = -x² - 2
Explanation:
Parent function: x²
Graph g(x) is inverted so -x² and then shifted down 2 units so -x² - 2
Hence: g(x) = -x² - 2
Solve the equation. 5/x+3 + 4/x+2 = 2
Please explain this step by step
Answer:
Step-by-step explanation: If I have to Write In Slope- Intercept form, This Is how I explain it.
Since x = −3 is a vertical line, there is no y-intercept and the slope is undefined.
Slope: Undefined
y-intercept: No y-intercept
Students are given 3 minutes to complete each multiple-choice question on a test and 8 minutes for each free-response question. There are 15 questions on the test and the students have been given 55 minutes to complete it
Let
x------> the number of multiple choice question
y------> the number of free response question
we know that
-----> equation A
-----> equation B
Substitute equation B in equation A
Find the value of x
therefore
the answer is
the number of multiple choice question are
the number of free response question are
PLEASE HELP URGENT
A triangle has one side that is 5 units long, one 25° angle, and one 90° angle. Complete the two diagrams to create two different triangles with these measurements. Label the 90° angle in each diagram.
Answer:
see attached
Step-by-step explanation:
The two different triangles can be formed by placing the 90° angle adjacent to, or opposite the given side.
__
In the attached diagram, the two triangles are ABC and ABD. The right angles are at vertex C and vertex B, respectively.
A cone has a height of 15 feet and a radius of 11 feet. What is its volume?
Use ≈ 3.14 and round your answer to the nearest hundredth.
cubic feet?
Answer: 1899.70 cu ft
Step-by-step explanation:
Cone volume = 1/3 TT r^2 h
V= (1/3) (3.14) (11^2) (15)
V=1899.70 cu ft
The function ((-1,2),(9-1),(-8,5),(2,-8)(?,9)has the same set for its domain and range.
Find the missing value from the function. The missing value is _____.
=======================================================
Explanation:
The range is {2, -1, 5, -8, 9} which sorts to {-8, -1, 2, 5, 9}. This is the set of y coordinates of the points given. Each point is of the form (x,y).
Since the domain and range are the same set in this case, we must have those value mentioned as x coordinates of the points.
The domain is the set of x coordinates of the points given.
The values -8, -1, 2, and 9 are already used as x coordinates. The only thing missing is the 5.
The scatterplot shows no correlation, a positive correlation, a negative correlation. Because the values of y decrease, increase, stay the same. As the values of x increase
The scatterplot shows no correlation, a positive correlation, a negative correlation. Because the values of y stay the same, increase, decrease, as the values of x increase.
What is scatterplot?Scatterplot shows relation between two quantitative variables measured for the same element.
The negative correlation shows that y decreases when x increases.
The positive correlation shows that y increases when x decreases.
No correlation shows shows that there is no change in the value of y when x stay same.
Thus, The scatterplot shows no correlation, a positive correlation, a negative correlation. Because the values of y stay the same, increase, decrease, as the values of x increase.
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The graph shows two lines, A and B.
A graph is shown with x- and y-axes labeled from 0 to 6 at increments of 1. A straight line labeled A joins the ordered pair 2, 6 with the ordered pair 5, 0. Another straight line labeled B joins the ordered pair 0, 2 with the ordered pair 6, 6.
Part A: How many solutions does the pair of equations for lines A and B have? Explain your answer. (5 points)
Part B: What is the solution to the equations of lines A and B? Explain your answer. (5 points)
Answer:
A. one solution; lines intersect at one point
B. (x, y) = (3, 4)
Step-by-step explanation:
A linear equation will be represented on a graph by a straight line. A pair of equations will graph as a pair of lines. If the lines are not coincident or parallel, they will intersect at exactly one point. That point is the solution to the pair of equations.
__
A.A graph of the given points quickly shows the lines are not parallel or coincident. Each point on line A satisfies the equation that describes the line. Each point on line B satisfies the equation that describes line B. The point of intersection is a point that satisfies both equations.
A pair of distinct non-parallel lines can have at most one point of intersection. Hence there can only be one solution to the pair of equations.
__
B.The graph shows the solution to the equations for lines A and B is ...
(x, y) = (3, 4)
List at least three shapes, other than pentagons, in the table. If the shape is a polygon, indicate whether the shape is regular or irregular. Find the number of times that will the shape will map back on to itself as the shape rotates 360° about its center. Also note how many degrees the shape has rotated each time it maps back onto itself. Use GeoGebra to guide you in this exercise, if you wish.
The number of times that the shape will map back onto itself as the shape rotates 360° about its center is 6.
What is the sum of all the exterior angles of a regular polygon?For a regular polygon of any number of sides, the sum of its exterior angle is 360° (full angle).
Regular polygons have all sides the same and that apothem bisects the side in two parts, (provable by symmetry).
A regular polygon with n sides can map onto itself by n times
It will rotate 360°/n about its center every time and will map onto itself
Examples are in the attached table as well as the solution.
The number of times that the shape will map back onto itself as the shape rotates 360° about its center is 6.
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By percent, how much greater do sales of horror books appear to be than sales of cookbooks? how much greater are they in reality? a. there appear to be 50% more sales of horror books than cookbooks, but there are really only about 20% more. b. there appear to be 40% more sales of horror books than cookbooks, but there are really only about 5% more. c. there appear to be 20% more sales of horror books than cookbooks, but there are really only about 10% more. d. there appear to be 25% more sales of horror books than cookbooks, but there are really only about 2% more.
There appear to be 25% more sales of horror books than cookbooks, but there are really only about 2% more.
How much greater is the sale of horror books compared to cookbooks?
Percentage is a measure of frequency that expresses the ratio of two numbers as a value out of 100.
Percentage of horror books to comedy books = [(number of horror books sold / number of comedy book) - 1 ] x 100
[(55 / 54) - 1] x 100 = 2%
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If you have (4-9x)+23/9 what is the value of x
A 17
B 23
C 5
D 2
Step-by-step explanation:
Do you have a final product for the equation.
I am unable to find the missing part of the peoblome without a total to sum up to.
Sorry...
9/12 devided by 2/12
[tex]\textbf{Given that,}\\\\~~~\dfrac 9{12} \div \dfrac{2}{12}\\\\\textbf{We can rewrite it as}\\\\~~~~\dfrac{9}{12} \times \dfrac 1{\tfrac{2}{12}}\\\\\\=\dfrac {9}{12} \times \dfrac{12}2\\\\\\=\dfrac 92\\\\\\=4.5\\\\\textbf{Hence the quotient is}~ \dfrac 92 ~ \textbf{or}~ 4.5.[/tex]
algebra 2 pls help lol
Answer:
right -2down 2reflected in the y-axiscompressed by a factor of 1/2Step-by-step explanation:
The order of the transformations affects the values used for translation. Here, the translation is described before the reflection and compression.
__
reflectionThe grayed-out graph of the square root function opens to the right. The darker graph of the transformed function opens to the left, so the reflection is across the y-axis.
__
compressionThe square root function goes through the point (1, 1), that is, 1 unit right of the vertex, and 1 unit up. The transformed function goes through the point (1, -1/2), that is, 1 unit left of the vertex and 1/2 unit up. The vertical height of 1 unit on the original graph is compressed to a height of 1/2 unit on the transformed graph. Vertical compression is by a factor of 1/2.
__
translationThe description of the transformation gives the translation before the reflection and compression. So, to find where the uncompressed and unreflected vertex is, we need to reverse those transformations.
Expanding the transformed square root graph by a factor of 2 will undo its compression by a factor of 1/2. That will move the vertex from (2, -1) to (2, 2×(-1)) = (2, -2).
Reflecting this expanded function back across the y-axis will undo the original reflection. That moves the vertex from (2, -2) to (-2, -2). This is where the original translation left the function's vertex before the reflection and compression moved it to the location shown.
The translation is right -2 and down 2.
_____
Additional comment
It is possible that you will get pushback on these answers. If so, you should have your teacher demonstrate the transformations in the order described.
__
In the order described, we have ...
[tex]f(x)=\sqrt{x}\\\\f_1(x)=\sqrt{x-(-2)}=\sqrt{x+2}\qquad\text{translation right -2}\\\\f_2(x)=\sqrt{x+2}-2\qquad\text{translation down 2}\\\\f_3(x)=\sqrt{-x+2}-2\qquad\text{reflection in the y-axis}\\\\g(x)=\dfrac{1}{2}(\sqrt{-x+2}-2)\qquad\text{compression vertically by $\dfrac{1}{2}$}[/tex]
The attached graph shows the result of this sequence of transformations.
__
If the reflection and compression were done before the translation, then the translation would be 2 right and 1 down.
Sally has only 10-pence and 50 pence coins in her purse. She has 21 coins altogether with a total value of $5.30. Calculate how many of each coin type does she have?
Answer:
It is 13 and 8. Look down below and you will know the answer is right:
Step-by-step explanation:
10x+50(21-x)=530
10x+1050-50x=530
-40x=530-1050
-40x=-520
x=-520/-40
x=13
10 cent coin 13
50 cent coin 8
Hope it helps and mark me as brainliest everyone!!!
what is the simplified forom of i^14
Answer:
i4=1
Step-by-step explanation:
Explanation: Rewrite i14 as (i4)3×i2 . If i=√−1, then i2=−1 . From here (i2)2=(−1)2 , so i4=1
Given the equation 2(3x − 4) = 5x + 6, solve for the variable. Explain each step and justify your process.
Charlie solved a similar equation below. Is Charlie's solution correct? Explain why or why not.
4x − 3 = 2(x − 1)
4x − 3 = 2x + 2
2x − 3 = 2
2x = 6
x = 3 (10 points)
What does a plant cell have that an animal cell does not have? (select all that apply.) please answer me
Step-by-step explanation:
plant cells have a cell wall for to prepared their own food but animal cell haven't cell wall
plant cell have a large vacuoles but in plant cell it has Avery small vacuoles the two have the main difference between animal cell plant cell
Solve:
4(x + 7) = 5x = 2
Answer:
x = 26
Step-by-step explanation:
4(x+7) = 5x+2 1. (4) with x and 7
4x+28 = 5x+2 2. -5x on both sides
-x+28 = 2 3. -28 on both sides
-x = -26 4. /-1 on both sides
x = 26
lynn has 19 songs in her favourite playlist. hazel has eight more songs than lynn in hers. how many songs are in hazel's favourite playlist
can someone please help me find a line passing through the points (3,8) and (5,10)? I will reward branliest to the one who gets it correct.
Answer:
maybe the answer is x-y =-5
Question in the picture can anyone explain me how to convert mixed to improper??
1/6+2/4
The addition of the given fractions ( 1/6 + 2/4 ) to its simplest fraction form is 2/3.
What is the addition of the fraction?Given that;
1/6 + 2/4Addition = ?1/6 + 2/4
First, multiply the denominators then cross-multiply nominators and denominators.
1/6 + 2/4 = ( (6×2) + ( 4×1 ) ) / 6 × 4
1/6 + 2/4 = ( 12 + 4 ) / 24
1/6 + 2/4 = 16 / 24
We simplify
1/6 + 2/4 = 16 / 24 = 2/3
Therefore the addition of the given fractions ( 1/6 + 2/4 ) to its simplest fraction form is 2/3.
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Please I need help asap
Answer:
Option D
Step-by-step explanation:
[tex]\frac{2}{3}x=\frac{12}{5}[/tex]
[tex]x=(\frac{12}{5} )(\frac{3}{2} )=\frac{36}{10} =\frac{18}{5} =3\frac{3}{5}[/tex]
Hope this helps
Which of the following statements is true about rays?
OA. They are zero-dimensional objects
B. They are one-dimensional objects,
OC. They are three-dimensional objects
OD. They are two-dimensional objects
Answer: B. They are one-dimensional objects
This applies to line segments and lines as well.
One dimensional objects are such that they only have one component to them when it comes to determining a location on that object.
For instance, a number line is one dimensional because we only need one number to determine where a location is.
A ray is effectively a half-line, or put another way, a line is the combination of two rays glued together. Basically a ray points forever in one direction only, as a line will point in two opposite directions.
Extra side notes:
I can't think of any zero dimensional objects off the top of my head.3D space is a 3 dimensional object because we need 3 coordinates to determine an exact location.A flat 2D plane is a 2 dimensional object because we need 2 coordinates to determine a location.Answer: B
Step-by-step explanation:
If α and β are the zeros of the quadratic polynomial f(x) = 6x²+x-2, then the value of
1. α²+β²
2. 1/α+1/β
Can Someone Answer This Quick Thank You<3
Answer:
Given function:
[tex]f(x)=6x^2+x-2[/tex]
To find the zeros of the function, set the function to zero and factor:
[tex]\implies 6x^2+x-2=0[/tex]
[tex]\implies 6x^2+4x-3x-2=0[/tex]
[tex]\implies 2x(3x+2)-1(3x+2)=0[/tex]
[tex]\implies (2x-1)(3x+2)=0[/tex]
Therefore, the zeros are:
[tex]\implies (2x-1)=0 \implies x=\dfrac{1}{2}[/tex]
[tex]\implies (3x+2)=0 \implies x=-\dfrac{2}{3}[/tex]
If α and β are the zeros of the function:
[tex]\textsf{Let } \alpha=\dfrac{1}{2}[/tex][tex]\textsf{Let } \beta=-\dfrac{2}{3}[/tex]Question 1
[tex]\begin{aligned}\implies \alpha^2+\beta^2 & =\left(\dfrac{1}{2}\right)^2+\left(-\dfrac{2}{3}\right)^2\\\\& = \dfrac{1}{4}+\dfrac{4}{9}\\\\& = \dfrac{9}{36}+\dfrac{16}{36}\\\\& = \dfrac{25}{36}\end{aligned}[/tex]
Question 2
[tex]\begin{aligned}\implies \dfrac{1}{\alpha}+\dfrac{1}{\beta} & = \dfrac{1}{\frac{1}{2}}+\dfrac{1}{-\frac{2}{3}}\\\\& = 1 \times \dfrac{2}{1}+1 \times -\dfrac{3}{2}\\\\& = 2 - \dfrac{3}{2}\\\\& = \dfrac{1}{2}\end{aligned}[/tex]
Answer:
[tex]1) ~\alpha^2+\beta^2 = \dfrac{25}{36}\\\\\\2)~\dfrac 1 {\alpha} + \dfrac 1{\beta} = \dfrac 12[/tex]
Step-by-step explanation:
[tex]\text{Given that,}\\\\f(x) = 6x^2 +x -2~ \text{and the roots are}~ \alpha, \beta\\\\\text{Now,}\\\\\alpha + \beta = -\dfrac{b}{a} = -\dfrac{1}{6}~~~~~;[\text{Compare with the standard form}~ ax^2 +b x + c = 0]\\\\\alpha \beta = \dfrac ca = -\dfrac2 6 = - \dfrac 13\\\\\\\textbf{1)}\\\\\alpha^2 +\beta^2\\\\\\=\left(\alpha +\beta \right)^2 - 2 \alpha \beta \\\\\\=\left( -\dfrac 16 \right)^2 -2 \left(- \dfrac 13 \right)\\\\\\=\dfrac{1}{36}+\dfrac 23\\\\\\=\dfrac{25}{36}[/tex]
[tex]\textbf{2)}\\\\\dfrac 1{\alpha} + \dfrac 1{\beta} \\\\\\=\dfrac{\alpha + \beta}{\alpha \beta }\\\\\\=\dfrac{-\tfrac16}{-\tfrac 13}\\\\\\=\dfrac{1}{6} \times 3\\\\\\=\dfrac{1}{2}[/tex]
Factor the expressions below using the greatest
common factor:
a. 4m³ - 32m
b. 63x¹2 - 35x6
C. -28v² - 8v - 36
Answer:
a=4m
b=63x¹2 - 35x6 maybe you formatted/typed this wrong, but this is impossible..?
c=4
28. Which of the following is the correct measure of
angle ABC?
A. 30°
B. 88°
C. 132°
D 44°
Answer:
D) 44°Step-by-step explanation:
According to the diagram we observe:
∠ACD is the exterior angle of triangle ABC.As we know the exterior angle is same as the some of the remote interior angles.
It can be shown as:
m∠ACD = m∠CAB + m∠CBA
Substitute the values and solve for x:
5x - 18 = 3x - 2 + x + 145x - 18 = 4x + 125x - 4x = 12 + 18x = 30Find the measure of m∠ABC:
x + 14 = 30 + 14 = 44The matching answer choice is D.
Please answer this question, i request
If cot θ = 7/8 , evaluate :-
(1 + sin θ)(1 – sin θ)/(1 + cos θ)(1 - cos θ)
[tex]{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}[/tex]
[tex] \star \: \tt \cot \theta = \dfrac{7}{8} [/tex]
[tex] {\large{\textsf{\textbf{\underline{\underline{To \: Evaluate :}}}}}}[/tex]
[tex] \star \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) }[/tex]
[tex]{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}[/tex]
Consider a [tex]\triangle[/tex] ABC right angled at C and [tex]\sf \angle \: B = \theta [/tex]
Then,
‣ Base [B] = BC
‣ Perpendicular [P] = AC
‣ Hypotenuse [H] = AB
[tex] \therefore \tt \cot \theta = \dfrac{Base}{ Perpendicular} = \dfrac{BC}{AC} = \dfrac{7}{8}[/tex]
Let,
Base = 7k and Perpendicular = 8k, where k is any positive integer
In [tex]\triangle[/tex] ABC, H² = B² + P² by Pythagoras theorem
[tex] \longrightarrow \tt {AB}^{2} = {BC}^{2} + {AC}^{2} [/tex]
[tex] \longrightarrow \tt {AB}^{2} = {(7k)}^{2} + {(8k)}^{2} [/tex]
[tex]\longrightarrow \tt {AB}^{2} = 49{k}^{2} + 64{k}^{2} [/tex]
[tex]\longrightarrow \tt {AB}^{2} = 113{k}^{2} [/tex]
[tex]\longrightarrow \tt AB = \sqrt{113 {k}^{2} } [/tex]
[tex]\longrightarrow \tt AB = \red{ \sqrt{113} \: k}[/tex]
Calculating Sin [tex]\sf \theta [/tex]
[tex] \longrightarrow \tt \sin \theta = \dfrac{Perpendicular}{Hypotenuse}[/tex]
[tex] \longrightarrow \tt \sin \theta = \dfrac{AC}{AB}[/tex]
[tex]\longrightarrow \tt \sin \theta = \dfrac{8 \cancel{k}}{ \sqrt{113} \: \cancel{ k } }[/tex]
[tex]\longrightarrow \tt \sin \theta = \purple{ \dfrac{8}{ \sqrt{113} } }[/tex]
Calculating Cos [tex]\sf \theta [/tex]
[tex] \longrightarrow \tt \cos \theta = \dfrac{Base}{Hypotenuse}[/tex]
[tex] \longrightarrow \tt \cos \theta = \dfrac{BC}{ AB} [/tex]
[tex] \longrightarrow \tt \cos \theta = \dfrac{7 \cancel{k}}{ \sqrt{113} \: \cancel{k } }[/tex]
[tex]\longrightarrow \tt \cos \theta = \purple{ \dfrac{7}{ \sqrt{113} } }[/tex]
Solving the given expression :-
[tex] \longrightarrow \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) } [/tex]
Putting,
• Sin [tex]\sf \theta [/tex] = [tex]\dfrac{8}{ \sqrt{113} }[/tex]
• Cos [tex]\sf \theta [/tex] = [tex]\dfrac{7}{ \sqrt{113} }[/tex]
[tex] \longrightarrow \: \tt \dfrac{ \bigg(1 + \dfrac{8}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{8}{ \sqrt{133}} \bigg) }{\bigg(1 + \dfrac{7}{ \sqrt{133}} \bigg) \bigg(1 - \dfrac{7}{ \sqrt{133}} \bigg)} [/tex]
Using (a + b ) (a - b ) = a² - b²
[tex]\longrightarrow \: \tt \dfrac{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{8}{ \sqrt{133} } \bigg)}^{2} }{ { \bigg(1 \bigg)}^{2} - { \bigg( \dfrac{7}{ \sqrt{133} } \bigg)}^{2} } [/tex]
[tex]\longrightarrow \: \tt \dfrac{1 - \dfrac{64}{113} }{ 1 - \dfrac{49}{113} } [/tex]
[tex]\longrightarrow \: \tt \dfrac{ \dfrac{113 - 64}{113} }{ \dfrac{113 - 49}{113} } [/tex]
[tex]\longrightarrow \: \tt { \dfrac { \dfrac{49}{113} }{ \dfrac{64}{113} } }[/tex]
[tex]\longrightarrow \: \tt { \dfrac{49}{113} }÷{ \dfrac{64}{113} }[/tex]
[tex]\longrightarrow \: \tt \dfrac{49}{ \cancel{113}} \times \dfrac{ \cancel{113}}{64} [/tex]
[tex]\longrightarrow \: \tt \dfrac{49}{64} [/tex]
[tex]\qquad \: \therefore \: \tt \dfrac{(1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta) (1 - \cos \theta) } = \pink{\dfrac{49}{64} }[/tex]
[tex]\begin{gathered} {\underline{\rule{300pt}{4pt}}} \end{gathered} [/tex]
[tex] {\large{\textsf{\textbf{\underline{\underline{We \: know :}}}}}}[/tex]
✧ Basic Formulas of Trigonometry is given by :-
[tex]\begin{gathered}\begin{gathered}\boxed { \begin{array}{c c} \\ \bigstar \: \sf{ In \:a \:Right \:Angled \: Triangle :} \\ \\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}}\\\\ \sf{\star \cosec \theta = \dfrac{Hypotenuse}{Perpendicular}} \\\\ \sf{\star \sec \theta = \dfrac{Hypotenuse}{Base}}\\\\ \sf{\star \cot \theta = \dfrac{Base}{Perpendicular}} \end{array}}\\\end{gathered} \end{gathered}[/tex]
[tex]{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}[/tex]
✧ Figure in attachment
[tex]\begin{gathered} {\underline{\rule{200pt}{1pt}}} \end{gathered} [/tex]