11 5/8 +9 1/2 is my problem

faltersainse [42]

$\text{Hello there!}\\\\\text{A simple way to find the value of x is to find the scale factor}\\\\\text{When you flip the shapes to make it look exactly the same,}\\\text{you would see that the numbers are divisible to it}\\\\\text{Side 21 is congruent to side 3}\\\\\text{Divide 21 by 3 and you'll get 7 (1/7 sine we're going to a smaller shape)}\\\\\text{The x side is congruent to the 91 side}\\\\\text{Divide 91 by 7}\\\\91\div7=13\\\\\large\boxed{x=13}$

7 0

3 years ago

Read 2 more answers

Greg rented a truck for one day there was a base fee of $16.99 and there was additional charge of $.89 for each mile driven Greg

julia-pushkina [17]

Let m represent the number of mile Greg drove. His rental charge was
207.45 = 16.99 + 0.89m
Subtract 16.99 and divide by 0.89
190.46 = 0.89 m
214 = m

Greg drove the truck 214 miles.

7 0

3 years ago

Adding and subtracting the given polynomials

Usimov [2.4K]

You have to combine like terms, so the variable (x, y, s, d, c....) and the exponents must be the same in order to combine them.

For example:

x² + x³ Since they don't have the same exponent, you can't combine them

y² + 3y² = 4y²

23x + x = 24x

4. 2s² + 1 + s² - 2s + 1 You can rearrange it if it makes it easier

2s² + s² - 2s + 1 + 1 = 3s² - 2s + 2

5. 5t² - 2t - 1 - (3t² - 5t + 7) Distribute/multiply the - to (3t² - 5t + 7)

5t² - 2t - 1 - 3t² + 5t - 7 = 2t² + 3t - 8

Do the same for #9 and #10, and you should get:

9. 2k² + 5k - 9

10. 6y³ - 7y² - 6y - 12

3 0

3 years ago

Andrew claims the initial value and y - intercept are the same thing on a graph. Is he correct?

Agata [3.3K]

Answer:

We conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

Step-by-step explanation:

We know that the initial value on a graph is basically the out-put value y of the point where the line meets or crosses the y-axis.

In other words, the initial value is the y-value or output of the point at x = 0

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the initial value of the equation y = 2x+1 is: y = 1

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Thus, at y = 1, the line meets the y-axis.

Hence, the initial value of the line is: y = 1

Similarly, we know that the value of the y-intercept can be determined by setting x = 0 and determining the corresponding value of y.

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the y-intercept of y = 2x+1 is y = 1.

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Therefore, the y-intercept of y = 2x+1 is y = 1.

Conclusion:

Therefore, we conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

6 0

2 years ago

Is it possible for the two triangles described below to be similar? Two angles of one triangle measure 60° and 70°. Two angles o

Aleonysh [2.5K]

Answer:

no

Step-by-step explanation:

Similar triangles have angles with the same measures.

angles in the first triangle are 50°, 60°, 70°.

angles in the second triangle are 50°, 50°, 80°.

The angles do not have the same values, so the triangles cannot be similar.

_____

<em>Comment on the question</em>

When 2 of the three angles are different, the third angle cannot make the triangles similar. We don't even need to compute the third angle to answer this question.

5 0

3 years ago