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4 years ago

What is the surface area of a box if it is scaled up by a factor of 10

1 answer:

7 0

If we are scaling up the surface area by a factor of 10 means that the two dimensions of area are each being increased by 10 times the original size. Therefore, the surface area is increased by a factor of 10 x 10, or by a factor of 100.

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Trevor Benjamin is a salesperson who is paid a monthly salary of $500 plus 2% commission on sales. Write an equation that repres

Varvara68 [4.7K]

Let, the amount of sales = x
Then, commission from sale will be: x * 0.02

So, Equation would be: 500 + 0.02x

Hope this helps!

6 0

3 years ago

Read 2 more answers

Each side of a triangle is nine inches long. find the perimeter

ella [17]

Answer:

Step-by-step explanation:

sorry if its wrong i just did 9×9×9 and got 27

5 0

3 years ago

Justify your reasoning when solving the equation: 3(x+4)=8x-18

vampirchik [111]

Answer:

Step-by-step explanation:

3x plus 12= 8x -18

subtract twelve on both sides:

3x= 8x- 30

minus 8x on both sides:

-5= -30

divide:6

3 0

3 years ago

Adam works at a bakery. He uses 1/8 of his flour supply to make bread and 3/4 of his flour supply to make other baked goods. If

kykrilka [37]

Answer: B

he has 24 in his supply!

Step-by-step explanation:

7 0

3 years ago

The force of gravity on Mars is different than on Earth. The function of the same situation on Mars would be represented by the

sweet-ann [11.9K]

Answer:

If thrown up with the same speed, the ball will go highest in Mars, and also it would take the ball longest to reach the maximum and as well to return to the ground.

Step-by-step explanation:

Keep in mind that the gravity on Mars; surface is less (about just 38%) of the acceleration of gravity on Earth's surface. Then when we use the kinematic formulas:

$v=v_0+a\,*\,t\\y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2$

the acceleration (which by the way is a negative number since acts opposite the initial velocity and displacement when we throw an object up on either planet.

Therefore, throwing the ball straight up makes the time for when the object stops going up and starts coming down (at the maximum height the object gets) the following:

$v=v_0+a\,*\,t\\0=v_0-g\,*\,t\\t=\frac{v_0}{t}$

When we use this to replace the 't" in the displacement formula, we et:

$y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2\\y-y_0=v_0\,(\frac{v_0}{g} )-\frac{g}{2} \,(\frac{v_0}{g} )^2\\y-y_0=\frac{1}{2} \frac{v_0^2}{g}$

This tells us that the smaller the value of "g", the highest the ball will go (g is in the denominator so a small value makes the quotient larger)

And we can also answer the question about time, since given the same initial velocity $v_0$ , the smaller the value of "g", the larger the value for the time to reach the maximum, and similarly to reach the ground when coming back down, since the acceleration is smaller (will take longer in Mars to cover the same distance)

3 0

3 years ago