3. Find the two numbers whose difference is 8 and the sum of twice the first and 3 times the

Solve for m:<br> -5(m - 3) + 6 = 2(m - 5)

tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

-5(m - 3) + 6 = 2(m - 5)

<u>Expand the terms in the bracket</u>

That's

- 5m + 15 + 6 = 2m - 10

- 5m + 21 = 2m - 10

<u>Subtract 2m from both sides</u>

- 5m - 2m + 21 = 2m - 2m - 10

- 7m + 21 = - 10

<u>Subtract 21 from both sides</u>

That's

- 7m + 21 - 21 = - 10 - 31

- 7m = - 31

<u>Divide both sides by - 7</u>

We have the final answer as

Hope this helps you

4 0

3 years ago

What is the area of this figure?

KatRina [158]

Hi!

I have attached 2 images that should help you understand :)

First, look at the edits I made to the image you posted. I separated the shape into smaller shapes so that we can find the area of each individual one.

Let's start with the rectangle.

To find the area of a rectangle, multiply the width times the height.
10 · 4 = 40

Rectangle = 40cm

Next up, the red triangles.

I have included another image showing the triangles combined into rectangles. So we can find the area of the triangles just like we would rectangles!

(let me know if you don't understand how I found the width + height of the triangles)

5 · 10 = 50

Red triangles = 50cm

And finally, the green triangles.

8 · 7 = 56

Green triangles = 56cm

Add it all together and you get...

40 + 50 + 56 = 146

The answer to the question is 146cm.

Next time you are having trouble with something like this, picture the triangles as rectangles! :)

3 0

3 years ago

Y = -(x + 5)(x + 1)

frez [133]

Answer:

dffdfdfdfdfsdewered

Step-by-step explanation:

8 0

3 years ago

A circular garden has a circumference of

omeli [17]

This solution to this problem is predicated on the fact that the circumference is just: $2\pir$ . A straight line going through the center of the garden would actually be the diameter, which is well known to be two times the radius of the circle, so we can say that the circumference is just:

$d\pi$

So, solving for both the radius and the diameter gives us:

$d\pi=8xy^2\pi\\ d=8xy^2\\ \\ 2\pi r=8xy^2\pi\\ r=4xy^2$

So, the length of thes traight path that goes through the center of the guardain is just $8xy^2$ , and we can use the radius for the next part of the problem.

The area of a circle is $\pi r^2$ , which means we can just plug in the radius and find our area:

$A=\pi r^2 \implies \\ A=\pi (4xy^2)^2\implies \\ A= 16\pi x^2y^4$

So, we have found our area( $16 \pi x^2 y^4$ ) and the problem is done.

6 0

3 years ago

A skyrocket is launched from a 30–foot-high platform, with initial speed 300 ft/s. If the polynomial −16t2 + 300t + 30 gives the

Stella [2.4K]

Answer:

566 ft

Step-by-step explanation:

Given the polynomial equation

-16t² + 300t + 30 = 0

We're asked to find how high the rocket would rise in a 2 second fuse rise. This is a simple instruction that can be obtained by setting the unknown, t, in the equation to 2. Doing this, we have

-16 * 2² + 300 * 2 + 30

-16 * 4 + 600 + 30

-64 + 630 = 566 ft

Then, we can therefore conclude that the rocket will rise high up to a distance of 566 ft

8 0

3 years ago