can someone help me find the missing ones​

Answer:

The answer is [tex]x > 2[/tex]

Step-by-step explanation:

to solve this inequality:

[tex]5x+7 > 17[/tex]

Put "-7" on both sides

[tex]5x+10[/tex]

divide by 5 on both sides

your answer should be 2.

The value of y based on the values given ib the equilateral triangle will be 4.

From the information given, MN is one of the sides of the equilateral triangle. Therefore, side MR will be:

= 1/2 × 8 = 4

In the above, it should be noted that MR was calculated since it's half of the value of MN which is 8.

Here, value of y based on the values given ib the equilateral triangle will be 4.

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Answer:

Step-by-step explanation:

4

Step-by-step explanation:

that means that in 5 ml there are 250mg.

and that means that in 1 ml there are 250/5 = 50 mg.

and now we can find out the mg in 7 ml (by multiplying everything by 7) :

7 ml = 50 × 7 = 350 mg

Answer:

First box ⇒ 3.6 x 10^-1

Second box ⇒ 3.1 x 10^6

Third box ⇒ 5.3 x 10^-6

Fourth box ⇒ 4.2 x 10^-1

Step-by-step explanation:

I used a calculator and solved it :)

Answer:

first square = 3.6 x [tex]10^{-1}[/tex]

Step-by-step explanation:

They are all multiplying so there is no need for parenthesis

(3.8 * 10³) * (9.4 * [tex]10^{-5}[/tex])

3.8 * 10³ * 9.4 * [tex]10^{-5}[/tex]
35.72 * [tex]10^{-2}[/tex]

35.72 =~ 36

Everytime you elevate a number, it's the same as...
3 * 10² = 100 * 3 = 300

0.3 * 10³ = 1000 * 0.3 = 300

so then...

36 *      [tex]10^{-2}[/tex] = 3.6 * [tex]10^{-1}[/tex]

well, the total amount of technicians employed is 100% and that is "x" amount, now, we also know that 3% of of that is 10800, so

[tex]\begin{array}{ccll} amount&\%\\ \cline{1-2} x & 100\\ 10800& 3 \end{array} \implies \cfrac{x}{10800}~~=~~\cfrac{100}{3} \\\\\\ 3x=1080000\implies x=\cfrac{1080000}{3}\implies x=360000[/tex]

[tex]|\Omega|=48\\A=\{24,48\}\\|A|=2\\\\P(A)=\dfrac{2}{48}=\dfrac{1}{24}\approx4.2\%[/tex]

Photosynthesis is the process by which plants use sunlight, water, and carbon dioxide to create oxygen and energy in the form of sugar.

[tex]\pink{ \rule{1pt}{90000000pt}}\color{blue}{ \rule{1pt}{90000000pt}}\color{yellow}{ \rule{1pt}{90000000pt}}\purple{ \rule{1pt}{90000000pt}} \green{\rule{1pt}{90000000pt}}

[/tex]

Answer:

4500

Step-by-step explanation:

all of them were admitted.

Answer:

D is the right ans after solving eqn we get two value of x and they are 1 and 3

Answer: 10

Step-by-step explanation:

It appears that [tex]\overline{AC} \perp \overline{BC}[/tex], so to check if this is the case, we can use the slope formula.

[tex]m_{\overline{AC}}=\frac{3-1}{6-2}=\frac{1}{2}\\m_{\overline{BC}}=\frac{3-7}{6-4}=-2[/tex]

Since these slopes are negative reciprocals, we can see that this is the case. So, we can take [tex]\overline{AC}[/tex] as the base of the triangle and [tex]\overline{BC}[/tex] as the height of the triangle, and then use the formula [tex]A=\frac{1}{2}bh[/tex].

Using the distance formula,

[tex]AC=\sqrt{(6-2)^{2}+(3-1)^{2}}=\sqrt{16+4}=2\sqrt{5}\\BC=\sqrt{(4-6)^{2}+(7-3)^{2}}=\sqrt{4+16}=2\sqrt{5}[/tex]

So, the area is [tex]\frac{1}{2}(2\sqrt{5})(2\sqrt{5})=\boxed{10}[/tex]

Using the vertically opposite angles theorem and corresponding angles theorem, we have proven that alternate exterior angles are congruent

From the question, we are to prove that alternate exterior angles are congruent.

In the given diagram, examples of alternate exterior angles are

∠1 and ∠8

∠2 and ∠7

Now, we will prove that ∠1 = ∠8

In the diagram, we can observe that

∠1 = ∠4 (Vertically opposite angles theorem)

and

∠4 = ∠8 (Corresponding angles theorem)

Then,

By the substitution property of equality

∠1 = ∠8

Hence, alternate exterior angles are congruent

Here is the complete and correct question:

Consider parallel lines cut by a transversal. Parallel lines q and s are cut by transversal r. On line q where it intersects line r, 4 angles are created. Labeled clockwise, from uppercase left: angle 1, angle 2, angle 4, angle 3. On line s where it intersects line r, 4 angles are created. Labeled clockwise, from uppercase left: angle 5, angle 6, angle 8, angle 7.

Explain which theorems, definitions, or combinations of both can be used to prove that alternate exterior angles are congruent.

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Answer:

Step-by-step explanation:

Parallel lines q and q are cut by transversal r forming 4 angles at each intersection. At the intersection of lines r and q, clockwise from top left, the angles are: 1, 2, 4, 3. At the intersection of lines r and s, clockwise from top left, the angles are: 5, 6, 8, 7.

Use the diagram to complete the statements.

Angles 1 and 5 are

✔ congruent

because they are

✔ corresponding

angles.

Angles 4 and 6 are

✔ supplementary

because they are

✔ same-side interior

angles.

Answer:

This equation has no solutions. So your answer is the last one!

Step-by-step explanation:

Answer:

The correct answer is D. none.

Answer:

E. 22

Step-by-step explanation:

Hello!

We know that b>6, and we are asked to find the possible values for 3b + 3.

Let's directly substitute 6 for b:

Now, because b is greater than 6, and not equal to 6, any value greater than 21 will work.

The correct option is E. 22.

Answer:

14/29

Step-by-step explanation:

14/29 is the answer

Answer: 30

Step-by-step explanation:

(Assuming the line is a tangent)

Answer:

[tex]x > -1\\[/tex]

Step by step explanation:

[tex]~~~~~2(4x-3) > -3(3)+5x\\\\\implies 8x-6 > -9+5x\\\\\implies 8x > -9+6+5x\\\\\implies 8x > -3+5x\\\\\implies 8x -5x > -3\\\\\implies 3x > -3\\\\\implies x > -\dfrac 33\\\\\implies x > -1[/tex]

Step-by-step explanation:

[tex] log_{5}(x + 9) = log_{5}(x + 5) + 1[/tex]

[tex] \frac{ log(x + 9) }{ log(5) } = \frac{ log(x + 5) }{ log(5) } + 1[/tex]

[tex] log(x + 9) = log(x + 5) + log(5) [/tex]

[tex] log(x + 9) = log(5x + 25) [/tex]

[tex]x + 9 = 5x + 25[/tex]

[tex]x = - 4[/tex]

If we plug it in, it is true

Using translation concepts, the function that represents g(x) is given by:

[tex]g(x) = -4\left(\frac{1}{2}\right)^x[/tex]

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

Function f(x) is given by:

[tex]f(x) = 4\left(\frac{1}{2}\right)^x[/tex]

Function g(x) is a reflection of g(x) over the x-axis, that is, is is multiplied by -1, hence:

[tex]g(x) = -f(x) = -4\left(\frac{1}{2}\right)^x[/tex]

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Answer:

5.5

Step-by-step explanation:

The longest string length is 12 3/4 (12.75)

The two shortest string lengths are 3 and 4 1/4 (4.25)

combine the length of the tow shortest string by adding them together,

3 + 4.25 = 7.25

to find how much longer the longest string you subtract the two numbers,

12.75 - 7.25 = 5.5

The term that could be put in the blank is x^2y^2

The polynomial is given as:

8x^3y^2 - ___ + 3xy^2 - 4y^3

From the above expression, we can see that each term has a decreasing power of x from x to 0.

The missing power of x in the expression is 2

So, the term that completes the expression must have x^2

Hence, the term that could be put in the blank is x^2y^2

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Answer:

its a

Step-by-step explanation:

If we need to turn this into a mathematical expression:

Answer: (x + y) - 9

Step-by-step explanation:

    9 less than the sum of x and y

    (the sum of x and y) - 9

    (x + y) - 9

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So

Hence

Option A

Answer:

Option A

Step-by-step explanation:

y=4x+10

So

Put x ax 12°C

Hence

y=4(12)+10

y=48+10

y=58

Answer:

-8

Step-by-step explanation:

If 3x-4y=2, then for any number a we also have a(3x-4y)=a(2) by multiplication property of equality.

Therefore, for a=-4 we have -4(3x-4y)=-4(2) which means the value of -4(3x-4y) is -8.

The value of equation - 4 (3x - 4y) will be;

⇒ - 4 (3x - 4y) = - 8

To multiply means to add a number to itself a particular number of times. Multiplication can be viewed as a process of repeated addition.

Given that;

The true equation is,

⇒ 3x - 4y = 2

Now,

Since, The true equation is,

⇒ 3x - 4y = 2   .. (i)

Hence, The value of equation - 4 (3x - 4y) is,

⇒ - 4 (3x - 4y)

Substitute from (i), we get;

⇒ - 4 (3x - 4y) = - 4 × 2

                      = - 8

Thus, The value of equation - 4 (3x - 4y) = - 8

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Answer:

D

Step-by-step explanation:

using the sine ratio in the right triangle and the exact value

sin45° = [tex]\frac{\sqrt{2} }{2}[/tex] , then

sin45° = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{x}{4}[/tex] = [tex]\frac{\sqrt{2} }{2}[/tex] ( cross- multiply )

2x = 4[tex]\sqrt{2}[/tex] ( divide both sides by 2 )

x = 2[tex]\sqrt{2}[/tex]

Answer:

Step-by-step explanation:

Equations

Let the dimes be x

Let the quarters be y

x + y = 19                     (1)                   Multiply this equation by 0.1

0.1x + 0.25y = 3.10     (2)

0.1x + 0.1 y =    1.90     (3)                   Subtract 3 from 2

          0.15y  =  1.20                            Divide by 0.15

          0.15y/0.15 = 1.20/0.15

                 y = 8

x + y = 19

x + 8 = 19                                               Subtract 8 from both sides

x+8-8  = 19-8                                          Combine

x = 11

Answer

Carol has 8 quarters

She has 11 dimes

Answer:

x = 10

Step-by-step explanation:

rewrite this equation with x on the left and numbers on the right

2x - 3x = 5 - 15

-x = -10

x=10

give brainliest, please!

hope this helps :)

The value of x is x = 10

To solve a first degree equation, first: we add or subtract like terms. Finally, we split the terms of both equalities.

- In this equation, we're going to subtract the like terms -- and on the other side of the equation, we're going to also subtract the real terms.

15 + 2x = 5 + 3x

2x - 3x = 5 - 10

-1x = -10

x = 10

So, the correct value is x = 10

The equation of the line about which the quadrilateral is reflected will be y = x.

It is the image of the point which is located in the opposite direction of a given point.

The rule of the refection is given will be

Firstly, the rule for reflecting a point about the line y = x is;

While reflecting on the line y = x, we get the reflected points by swapping the coordinates.

First, draw the line on the graph. Then reflect each vertex of quadrilateral across the line to produce quadrilateral.

Vertex Preimage Coordinates                  Vertex Image Coordinates

          A (-3,-2)                                                           A' (-2,-3)

          B (-2,-1)                                                            B' (-1,-2)

          C (-1,-2)                                                            C' (-2,-1)

          D (-1,-4)                                                            D' (-4,-1)

The diagram is given below.

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[tex]\\ \rm\Rrightarrow \dfrac{6}{x^2-3x}=\dfrac{12}{x}+\dfrac{1}{x-3}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{6}{x(x-3)}=\dfrac{12}{x}+\dfrac{1}{x-3}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{6}{x(x-3)}=\dfrac{12(x-3)+x}{x(x-3)}[/tex]

[tex]\\ \rm\Rrightarrow 6=12x-36+x[/tex]

[tex]\\ \rm\Rrightarrow 6=13x-36[/tex]

[tex]\\ \rm\Rrightarrow 13x=42[/tex]

[tex]\\ \rm\Rrightarrow x=\dfrac{42}{13}[/tex]