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3 years ago

14

Could you pls teach me this question

1 answer:

Alexandra [31]3 years ago

7 0

Answer:

if it's area the area would be the length of the diameter of the semi circle which is the height of the triangle multiplied by the base of the triangle to find the area of the triangle then to find the area of the circle would be pie multiplied by the diameter divided by two squared

$\pi \ \times 7 {}^{2}$

then the total area would be the area of the triangle plus the area of the half circle

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I need to know what the ansswer is to this we are doing pemdas

mamaluj [8]

Answer:

18

Step-by-step explanation:

10-4+6(4-2)

10-4+6(2)

10-4+12

6+12

=18

8 0

2 years ago

Find the root of each of the following equation x+8=24

kupik [55]

Answer:

4

Step-by-step explanation:

If you mean root of x, 24-8=16

square root of 16=4

4 0

2 years ago

Kim drives x miles a week. Terry drives two times more than Kim and Jerry drives three times more than Terry. Which expression r

liberstina [14]

The expression for the amount that Jerry drives is 6x. This can be concluded since 2*3=6. And how to represent it should be described as 6x.

4 0

3 years ago

Test the claim that the proportion of men who own cats is smaller than 70% at the 0.05 significance level. The null and alternat

Ghella [55]

Answer:

the null hypothesis would be: p = 70%/0.7

The alternative hypothesis would be: p < 0.7

Step-by-step explanation:

The null hypothesis is most of the time always the default statement while the alternative hypothesis is tested against the null and is its opposite.

In this case study the null hypothesis would be: the proportion of men who own cats is 70%: p = 0.7

The alternative hypothesis would be: the proportion of men who own cats is smaller than 70% : p < 0.7

3 0

3 years ago

Can someone help me with this problem

Alex_Xolod [135]

Answer:

<h2>A = 20</h2><h2>P = 6√10</h2>

Step-by-step explanation:

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

A(-3, 0) , B(3, 2)

$AB=\sqrt{(3-(-3))^2+(2-0)^2}=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}=\sqrt{4\cdot10}=\sqrt4\cdot\sqrt{10}=2\sqrt{10}$

A(-3, 0), D(-2, -3)

$AD=\sqrt{(-2-(-3))^2+(-3-0)^2}=\sqrt{1^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

AB = CD and AD = BC

The area of a rectangle:

$A=(AB)(AD)$

Substitute:

$A=(2\sqrt{10})(\sqrt{10})=(2)(10)=20$

The perimeter of a rectangle:

$P=2(AB+AD)$

Substitute:

$P=2(2\sqrt{10}+\sqrt{10})=2(3\sqrt{10})=6\sqrt{10}$

4 0

2 years ago