The area of a circle is 25pieft2. What is the circumference, in feet? Express your answer in terms of pie​

The equivalent resistance currents passing through each resistance  are: 1.8 A , 1.29A , 1A.

The total current passing through the circuit is  19.83 A.

The resistance 5Ω , 7Ω , 9Ω  which are are connected in parallel, then,

R1 =5Ω,

R 2=7Ω,

R3 =9Ω; supply voltage V=9 volt.

Effective resistance of parallel combination can be calculated as :

1/RP = 1/R1 + 1/R2 + 1/R3

1/RP= 1/5 + 1/7 +1/9

Rp = 0.45397Ω

Then the Current through  resistors

are R1,  I1 = V/R1 = 9/5 = 1.8A

R2,  I2 = V/R2 = 9/7 = 1.29A

R3,  I3 = V/R3 = 9/9 = 1A

Then Total current I = V/Rp = 9/ 0.45397 = 19.83 A

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In contrast to the signs of the real x-intercepts on the graph, we must remember the signs of the values in a quadratic's factored form. because the cartesian coordinate points are shifted.

Any function of the form [tex]\rm f(x) =ax^2+bx+c[/tex].where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic function.

When factoring by grouping is used, the replacement terms will both share the same middle-term sign. The factor signs will have one of each sign with the "stronger" number (the larger absolute value) matching the middle term if the last term is negative.

We can locate the function's x-intercepts or zeros with the aid of factored form.

Thus, in contrast to the signs of the real x-intercepts on the graph, we must remember the signs of the values in a quadratic's factored form. because the cartesian coordinate points are shifted.

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Answer: j = 9

Step-by-step explanation:

Slope = (y2-y1)/(x2-x1)

-1 = (3-(-5)) / (1-j)

-1 = (3+5) / (1-j)

-1 = 8 / (1-j)

Multiply both sides by 1 -j

= (1-j) = 8

= 1 - j = -8

= -j = -8 -1

j = 9

Given statement " If the measure of an angle is not 180, then it is not a straight angle" when related to conditional statement  "If an angle is a straight angle, then its measure is 180." then it represents the example of contrapositive.

As given in the question,

Given statement is equal to:

" If the measure of an angle is not 180, then it is not a straight angle"

And its conditional statement is given by:

"If an angle is a straight angle, then its measure is 180."

Explanation of the given options:

a. Counterexample : Concluding statement is against the given statement .

Here it is false.

b. Converse:  If it is not a straight angle then the measure of an angle is not 180.

Here it is false.

c. Inverse : If the measure of an angle is 180, then it is a straight angle.

False.

d. Contrapositive : "If an angle is a straight angle, then its measure is 180."

Here it is true.

Therefore,  for the given statement " If the measure of an angle is not 180, then it is not a straight angle" when related to conditional statement  "If an angle is a straight angle, then its measure is 180." then it represents the example of contrapositive.

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Answer: F = 1575

Step-by-step explanation: 1221 + 354 = 1575. So 1575 - 354 = 1221

Answer:

3(b-2)=24

Step-by-step explanation:

3b - 6 = 24

3b = 30

b= 10

For having a discriminant equal to or larger than zero we must have:

k ≤ 2 and k ≥ 10

For a quadratic equation of the form:

y = a*x^2 + b*x + c

The discriminant is:

D = b^2 - 4ac

And the equation has real roots only if the discriminant is equal to or larger than zero.

Here our equation is:

x^2 + (k - 2)*x + (2k - 4) = 0

The discriminant is:

D = (k - 2)^2 -4*1*(2k - 4)

D = k^2 -  4k + 4 - 8k + 16

D = k^2 -12k + 20

The solutions of :

k^2 -12k + 20 = 0

are:

k = (+12 ± √( (-12)^2 - 4*1*20))/(2)

k = (+12 ± 8)/2

The two values of k are:

k = 20/2 = 10

k = 4/2 = 2

The possible values of k are:

k ≤ 2 and k ≥ 10

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The relationship between pressure and temperature for a consisting of vapour and condensed phases is   [tex]l_{n}(p_{2}/p_{1}) = [ 1/T_{1}-1/T_{2}][(h_{g}-h_{f})/R][/tex].

The Clausius-Clapeyron equation is a differential equation which specifies the temperature dependence of pressure and vapour pressure. The Clausis-Clapeyron equation is named after Rudolf Clausius and Benoit Paul Emile Clapeyron.

The Clausius-Clapeyron equation at saturation temperature and pressure is

[tex]dP/dT[/tex]= [tex]h_{g} - h_{f}/T(v_{g}-v_{f} )[/tex]     →  1

Assuming that [tex]v_{g} > > > v_{f}[/tex] , we write

[tex]v _{g}[/tex] = RT/P   →   2  ( for both liquid and solid phases)

⇒     dP/dT   =   P([tex]h_{g} - h_{f}[/tex])/R[tex]T^{2}[/tex]

⇒     dP/P   =   dT/[tex]T^{2}[/tex]([tex]h_{g} - h_{f}[/tex]/R)   →  2

By integrating 2 , we get

                        [tex]l_{n}(p_{2}/p_{1}) = [ 1/T_{1}-1/T_{2}][(h_{g}-h_{f})/R][/tex]    →    3

Therefore, the equation to represent the relationship between pressure and temperature for a system consisting of vapour and condensed phases is equation 3  i.e.    [tex]l_{n}(p_{2}/p_{1}) = [ 1/T_{1}-1/T_{2}][(h_{g}-h_{f})/R][/tex]

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Answer:

7.81

Step-by-step explanation:

x1= 1

x2= 6

y1= -4

y2= 2

31.The first one just has one minor error which is not putting a square root.

after putting the square root on  [tex]\sqrt{61}[/tex] you will get an answer which is 7.81 units

the second one has an error in putting the correct values

32. the equation for distance is [tex]\sqrt{(x_2-x_1 )+(y_2-y_1)} }[/tex] however the values are inputted wrong as x2=6 and x1=1 but here x1 is taken as 2 which is the value of y2 and instead of inputting y2 as 2 it is written as 1 which is the value of x1

Comment if you still don't understand

Answer:

31.  Omission of the square root sign in step 1.

32.  Subtracting the y-value from the x-value in each parentheses in step 1.

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Distance between two points}\\\\$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}[/tex]

Given points:

The error was the omission of the square root sign in the first step of the calculation:

[tex]\textsf{Error}: \quad AB=(6-2)^2+(1-(-4))^2[/tex]

[tex]\textsf{Correction}: \quad AB=\sqrt{(6-2)^2+(1-(-4))^2}[/tex]

Correct calculation:

[tex]\begin{aligned}AB&=\sqrt{(6-1)^2+(2-(-4))^2}\\&=\sqrt{5^2+6^2}\\&=\sqrt{25+36}\\&=\sqrt{61}\\& \approx 7.8\end{aligned}[/tex]

The error was subtracting the y-value from the x-value in each parentheses in the first step of the calculation, rather than subtracting the x-values and the y-values separately:

[tex]\begin{aligned}\textsf{Error}: \quad AB&=\sqrt{(x_A-y_A)^2+(x_B-y_B)^2}\\ &=\sqrt{(6-2)^2+(1-(-4))^2}\end{aligned}[/tex]

[tex]\begin{aligned}\textsf{Correction}: \quad AB&=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}\\&=\sqrt{(6-1)^2+(2-(-4))^2\end{aligned}[/tex]

Correct calculation:

[tex]\begin{aligned}AB&=\sqrt{(6-1)^2+(2-(-4))^2}\\&=\sqrt{5^2+6^2}\\&=\sqrt{25+36}\\&=\sqrt{61}\\& \approx 7.8\end{aligned}[/tex]

Answer: Value of x is B equals 6

Step-by-step explanation:

Using the Side Angle Side Theorem, ∆KWL ≅ ∆ALW.

In the given question we have to

Prove: ∆KWL ≅ ∆ALW

Given: ∠KWL ≅ ∠WLK, ∠AWL ≅ ∠WLK

As given that

∠KWL ≅ ∠WLK

∠AWL ≅ ∠WLK

WL = WL (Diagonal)

Equal Parallel Sides

WK = AL

WA = KL

From Corresponding Angle Theorem

∠KWL=∠WLK and ∠AWL = ∠WLK

So ∠W=∠L

So from the Side Angle Side(SAS) Theorem

∆KWL ≅ ∆ALW.

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The congruence theorem that can be used to prove that the triangles are congruent is F. Not enough information is given to prove that the triangles are congruent

If all three corresponding sides are equal and all three corresponding angles are equal in measure, two triangles are said to be congruent.

According to the Angle-Side-Angle Theorem (ASA), if two angles and their included side are congruent to two angles and their included side, then these two triangles are congruent.

The hypotenuse is always the opposite side of the right angle. The legs of the triangle are the other two sides of a right triangle. Because a right triangle has two legs and a hypotenuse, it is also known as a hypotenuse-leg (or HL) triangle.

In this situation, enough information aren't given about the triangles. Therefore, the correct option is F.

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Answer:

[tex]y=4x-6[/tex]

Step-by-step explanation:

The slope is [tex]\frac{-2-2}{1-2}=4[/tex].

[tex]y-2=4(x-2) \\ \\ y-2=4x-8 \\ \\ y=4x-6[/tex]

Answer:

yes

Step-by-step explanation:

as long as one input does not have two different outputs it is a function

The equation for the total cost of the dog tags will be 348 + 2t.

The equation to show the revenue would be Revenue = 5t

The number of dog tags that Logan needs to sell to break even is 116 dog tags.

Assuming that the number of dog tags sold by Logan to raise money is represented by t, the total cost would be:

= Flat fee charge + Cost per tag x Number of tags

= 348 + 2t

The total revenue would be:

= Number of dog tags x Selling price per dog tag

= 5t

The breakeven point would be the point where total profit would be zero and so can be shown as:

0 = Sales - total cost

0 = 5t - (348 + 2t)

0 = 5t - 348 - 2t

3t = 348

t = 348 / 3

t = 116 tags

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In this question, line1 and line2 are parallel to each other by finding the slope.

What is the slope of line?

Given two line-side coordinates, use the slope formula to determine the slope of the line. The slope is defined as the ratio of the change in the y values to the change in the x values using the formula m=(y2-y1)/(x2-x1). x1 and y1 are represented by the first point's coordinates. The second points are located at x2, y2, and their locations. Whichever point you choose to classify as the first and which as the second doesn't matter.

We know that formula two find the slope of the line joining two points which is

Given four lines are

Line 1: y=1/2x -6

(Line 2: x-2y=-1

(Line 3: y=4x+6

(Line 4: y+3=1/4 (x-6)

We have to find out the slope of the corresponding lines such that

the slope of line 1 = 1/2

the slope of line 2 = 1/2

the slope of line 3 = 4

the slope of line 4 = 14

So the line has the same slope they are parallel to each other.

Hence in this question, line1 and line2 are parallel to each other.

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11 dummy variables will be needed.

A dummy variable is a numerical variable used in regression analysis to represent subgroups of the sample in your study. In research design, a dummy variable is often used to distinguish different treatment groups.

we know that,

dummy variable formula = K- 1,

here, K = 12 (months in a year)

this implies, K - 1 = 12 - 1 = 11

Therefore, no. of dummy variables needed = 11.

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the slope of the given point is m=−3/4

Answer:-3/4

Step-by-step explanation:

to find slope using 2 points the equation is ...

y2-y1

x2-x1

Check the picture below.

[tex]\stackrel{\measuredangle MNQ}{4x+16}\implies 4(14)+16\implies \stackrel{\measuredangle MNQ}{\text{\LARGE 72}}[/tex]

The domain of the relationship graphed, Domain: {x|x € N} is from -4 and ended at 4.

How to identify domain and range from co-ordinates?

If x,y be a co-ordinate pair then x is the domain and y is the range .

The domain of a function is the set of all possible inputs for the function.

The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain.

Here

x starts from -4 and ended at 4.

domain is the set of natural numbers

Hence,  the domain of the relationship graphed, Domain: {x|x € N} is from -4 and ended at 4.

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Step-by-step explanation:

Enter a problem...

Pre-Algebra Examples

Popular Problems Pre-Algebra Solve for x 12+2x-x=9x+6

12

+2

x−x=

9x

+6Subtract

x from

2x

12+x

=9x+6

Move all terms containing

x

to the left side of the equation.

Tap for more steps...

12−8x=6

to the right side of the equation.

-8x=−6

Divide each term in

−8x=−6 by −8

and simplify.

The result can be shown in multiple forms.

Exact Form:

x=3/4

Decimal Form:0.75

The trigonometric function with its right triangle definition is matched as below:

(a) sine

opposite/hypotenuse

(b) cosine

adjacent/hypotenuse

(c) tangent

opposite/adjacent

(d) cosecant

hypotenuse/opposite

(e) secant

hypotenuse/adjacent

(f) cotangent

adjacent/opposite

We Know That,

i) sine =opposite side of the triangle /hypotenuse side of the triangle

ii) cosine=adjacent side of the triangle/hypotenuse side of the triangle

iii) tangent=opposite side of the triangle/adjacent side of the triangle

iv) cosecant=hypotenuse side of the triangle/opposite side of the triangle

v) secant=hypotenuse side of the triangle/adjacent side of the triangle

vi) cotangent=adjacent side of the triangle/opposite side of the triangle.

Therefore, each trigonometric function is matched with its right triangle definition

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Answer:a

Step-by-step explanation:

Answer

-117/640

Step-by-step explanation:

ba + cd

b= -7/16

a= 7/8

c= 08.(4/5)

d= 1/4

Hence it can be calculated as follows

-7/16×7/8 + 4/5×1/4

-49/128 + 1/5

-246+128/640

= -117/640

Answer:

Step-by-step explanation:

this is all about your algebra skillz,     learn it , know it , live it :P

anyway

I'll make this very detailed , so you see the process.  but lean it

3n+8 = 5n-4

begin using reverse operation order,   note you need to know  P.E.M.D.A.S  ,   Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.   starting from the "Parenthesis,   and the working down thur each operation is forward operation order.  I bet nobody has told you this before, huh. So to do the algebra on the equation above use reverse operation order.

so, I'm going to move the -4 , first

3n+8+4 = 5n-4+4    ( do the same thing on both sides of the equal sign)

3n+12 = 5n

now I'm going to move the 3n

3n+12 -3n = 5n -3n  ( do the same on both sides again)

12 = 2n

now i'm moving up to Division,  b/c it's next on our order of operations,  i'll divide the 2 away from the n

12/2 = 2n / 2   ( same on both side again)

6 = n

now we have our answer for n,  nice :P  try this at home, school,  or at your local coffee shop  :P  

The completed-square (vertex) form 2x² + 12x + 1 is 2(x+3)^2−17. Completing the Square is a technique to find maximum or minimum values of quadratic functions.

Completing the square is a highly helpful technique or way to change a quadratic equation from also known as the "standard form," to its "vertex form,".

For some combinations of h and k, completing the square is a method for transforming a quadratic polynomial. In other words, the quadratic expression is completed by inserting a perfect square trinomial.

Finding the maximum or lowest values of quadratic functions can be done using the square method, sometimes known as "completing the square."

Use the form ax^2 + bx + c, to find the values of a, b, and c.

Consider the vertex form of a parabola.

a(x+d)^2+e

Find the value of d using the formula

d = b/2a.

d = 3

Find the value of e using the formula

e = c− b^2 / 4a.

e = −17

Substitute the values of a, d, and e into the vertex form 2(x+3)^2−17.

2(x+3)^2−17

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The slope of the line is -32, and the y-intercept is 480 if the equation of the line is y = -32x + 480.

A straight line is a combination of endless points joined on both sides of the point.

The slope 'm' of any straight line is given by:

A linear equation has the form y = mx + b in the slope-intercept format. X and Y are the variables in the equation. The values m and b represent the line's slope (m) and the value of y when x is 0.

[tex]\rm m =\dfrac{y_2-y_1}{x_2-x_1}[/tex]

The equation of the line uses two points:

(0, 480) and (15, 0)

y = (-480/15)[x - 15]

y = -32(x - 15)

y = -32x + 480

y = mx + c

m = -32

c = 480

Thus, the slope of the line is -32, and the y-intercept is 480 if the equation of the line is y = -32x + 480.

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The triangles ΔABC and ΔCDE are congruent to each other using the SSS rule.

We are given two triangles. The vertices of the first triangle are A, B, and C. The vertices of the second triangle are C, D, and E. We need to prove that the triangles are congruent to each other.

The length of side AB is equal to that of side CD. The length of side BC is equal to that of side DE. We know that C is the midpoint of line segment AE. This means that the length of AC is equal to the length of CE.

All three corresponding sides of the two triangles are equal. Hence, the triangles are congruent to each other using the SSS rule for congruency.

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Answer to part (A) is y = 42x+9

Answer to part (B) is 98

========================================================

Explanation:

Part (A)

Let's plug x = 0 into the 1st derivative of f(x)

[tex]f'(\text{x}) = \cos(\pi \text{x}) + \text{x}^5 + 6\\\\f'(0) = \cos(\pi *0) + (0)^5 + 6\\\\f'(0) = 1 + 0 + 6\\\\f'(0) = 7\\\\[/tex]

We'll use that later in the steps below, which show computing the derivative value of h(x) at x = 0.

[tex]h(\text{x}) = ( f(\text{x}) )^2\\\\h'(\text{x}) = 2( f(\text{x}) )*f'(\text{x}) \ \ \text{ ... chain rule}\\\\h'(0) = 2( f(0) )*f'(0)\\\\h'(0) = 2( 3 )*7\\\\h'(0) = 42\\\\[/tex]

This is the slope of the tangent line to h(x) at x = 0.

Now plug x = 0 into the h(x) function itself, without any derivatives applied.

[tex]h(\text{x}) = ( f(\text{x}) )^2\\\\h(0) = ( f(0) )^2\\\\h(0) = ( 3 )^2\\\\h(0) = 9\\\\[/tex]

This is the y intercept of the line, i.e. the b value.

We found that

We go from y = mx+b to y = 42x+9 as the equation of the tangent line.

===================================================

Part (B)

In the previous part, we already calculated the first derivative. Differentiate that with respect to x to get the second derivative.

[tex]h'(\text{x}) = 2( f(\text{x}) )*f'(\text{x})\\\\h''(\text{x}) = \frac{d}{dx}\left[h'(x)\right]\\\\h''(\text{x}) = \frac{d}{dx}\left[2( f(\text{x}) )*f'(\text{x})\right]\\\\h''(\text{x}) = \frac{d}{dx}\left[2( f(\text{x}) )\right]*f'(\text{x})+2*f(\text{x})*\frac{d}{dx}\left[f'(\text{x})\right] \ \ \text{ ... product rule}\\\\h''(\text{x}) = 2f'(\text{x})*f'(\text{x})+2*f(\text{x})*f''(\text{x})\\\\h''(\text{x}) = 2\left(f'(\text{x})\right)^2+2*f(\text{x})*f''(\text{x})\\\\[/tex]

The second derivative is useful to determine where the function is concave up or concave down. And also to determine points of inflection.

The h''(x) function involves f''(x), so we'll need to find the second derivative of the f(x) function.

[tex]f'(\text{x}) = \cos(\pi \text{x}) + \text{x}^5 + 6\\\\f''(\text{x}) = \frac{d}{dx}\left[\cos(\pi \text{x}) + \text{x}^5 + 6\right]\\\\f''(\text{x}) = -\pi\sin(\pi \text{x}) + 5\text{x}^4\\\\[/tex]

Then plug in x = 0

[tex]f''(\text{x}) = -\pi\sin(\pi \text{x}) + 5\text{x}^4\\\\f''(0) = -\pi\sin(\pi *0) + 5(0)^4\\\\f''(0) = 0\\\\[/tex]

We have enough info to find h''(0) finally.

[tex]h''(\text{x}) = 2\left(f'(\text{x})\right)^2+2*f(\text{x})*f''(\text{x})\\\\h''(0) = 2\left(f'(0)\right)^2+2*f(0)*f''(0)\\\\h''(0) = 2\left(7\right)^2+2*3*0\\\\h''(0) = 98\\\\[/tex]

Side notes: