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Nadusha1986 [10]

3 years ago

9

Help me find the scale factor

1 answer:

atroni [7]3 years ago

3 0

The scale factor for the first one is 2.25
And for the second one it’s 7
I hope this helps :)

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11. please help me outt ill mark you brainlist !

alex41 [277]

I’m not positive, but I think the answer is C because when you look at the big triangle, it looks like you divide each coordinate by 3 (or multiply it by 1/3) to get the small triangle.

e.g. 6 * 1/3 = 2, 3 * 1/3 = 1

8 0

3 years ago

What is the mean of this data set? {6, 11, 5, 2, 7} Enter your answer as a decimal in the box.

Elenna [48]

Answer:

The mean is 6.2.

Step-by-step explanation:

The "mean" is the same thing as the "average". Essentially, the question is asking for the average of the numbers.

So:

Add up all of the terms. [6 + 11 + 5 + 2 + 7] = 31

You find the average by dividing the sum of the terms (31) by the number of terms (5).

31/5 = 6.2

The mean is 6.2.

I hope this helped! :)

4 0

2 years ago

Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t

dmitriy555 [2]

Answer:

1. Find the difference between the areas.

<u>Area of the small rectangle</u>: $(x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14$

<u>Area of the big rectangle</u>: $(x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99$

The difference is: $11x+85$

$( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85$

2.

You can solve this question just by looking at the graph.

a) The height is 4 meters.

$f(d)=h=-2d^2+7d+4$

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

$f(0)=h=-2(0)^2+7(0)+4$

$h=4$

The height is 4 meters.

b) 9 meters.

For $d=1$

$f(1)=h=-2(1)^2+7(1)+4$

$f(1)=h=-2+7+4$

$h=9$

b) The ball travels 4 meters.

But to calculate it, it is when $h=0$

$0=-2d^2+7d+4$

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

3.

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

$V=\pi r^2h$

$V=\pi (2.5)^2 \cdot 5$

$V=31.25 \pi$

$V_{c} \approx 98.17 \text{ m}^3$

The volume of the sphere:

$V=\frac{4}{3} \pi r^2$

$V_{s} \approx 65.4 \text{ m}^3$

The volume of the capsule is approximately $163.57 \text{ m}^3$

3 0

3 years ago

Solve: 5x-4=6 please help meee

marta [7]

Answer:

x=2

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

How to graph this? Please asap

iVinArrow [24]

I hope this helps you

4 0

3 years ago