This is the other one for that one i couldn't hold it

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

I need help im not understanding how to do this

galina1969 [7]

Answer:

(c, b)

Step-by-step explanation:

Add the coordinates together and divide by 2

x coordinates: 2c + 0 = 2c, divided by 2 = c

y coordinates: 0 + 2b = 2b, divided by 2 = b

3 0

2 years ago

Read 2 more answers

Find the coefficient of the x^3 in the expansion of (2x-9)^5

il63 [147K]

Use the binomial theorem:

$\displaystyle (2x-9)^5 = \sum_{k=0}^5 \binom5k (2x)^{5-k}(-9)^k = \sum_{k=0}^5 \frac{5!}{k!(5-k)!} 2^5 \left(-\frac92\right)^k x^{5-k}$

The x ³ terms occurs for 5 - k = 3, or k = 2, and its coefficient would be

$\dfrac{5!}{2!(5-2)!} 2^5 \left(-\dfrac92\right)^2 = \boxed{6480}$

8 0

3 years ago

55 is 40% of what number

Oxana [17]

It is important to do the calculations very carefully. Otherwise there is nothing complicated about this problem.
Let us assume the unknown number = x
Then
40x/100 = 55
40x = 55 * 100
40x = 5500
x = 5500/40
= 550/4
= 137.5
So 55 is 40% of 137.50. I hope the procedure is clear to you and in future you can solve such problems on your own without requiring any help from outside.This is the simplest way of solving such problems.

5 0

3 years ago

Read 2 more answers

The sum of x and it’s opposite is always zero?

jeka94

Answer:

1) Let's consider the first case with the number 0 the oppose is also 0 and we have that 0-0=0 so then applies

2) Now let's consider any real number a no matter positive or negative we will have that:

$a-a =0$

Or in the other case:

$-a -(-a) =-a +a=0$

So then we can conclude that the expression is a general rule and is true

Step-by-step explanation:

For this case we can verify if the following expression is true or false:

The sum of x and it’s opposite is always zero?

If we want to proof this we need to show that for any number is true.

1) Let's consider the first case with the number 0 the oppose is also 0 and we have that 0-0=0 so then applies

2) Now let's consider any real number a no matter positive or negative we will have that:

$a-a =0$

Or in the other case:

$-a -(-a) =-a +a=0$

So then we can conclude that the expression is a general rule and is true

5 0

3 years ago