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3 years ago

(50 Points & Brainliest)

Mathematics

2 answers:

fomenos3 years ago

8 0

Answer:

D) There is no relationship between hours spent cycling and hours spent playing football.

Step-by-step explanation:

As we can see that the scatter plot dots are distributed very randomly and we cannot form any relation between both conditions. Rather to easily compare both conditions we need two plots with respect to time. So, we can compare the hours spent in cycling and hours spent in playing football.

So, the correct answer would be - There is no relationship between hours spent cycling and hours spent playing football.

givi [52]3 years ago

3 0

Answer:

Answer: There is no relationship between hours spent cycling and hours spent playing football. I did flvs and I got this question correct.

Step-by-step explanation:

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The product of 9 and m is 45

Jlenok [28]

9m = 45
m = 45/9 = 5.

4 0

3 years ago

Find the Area. Simplify your answer.

Vaselesa [24]

Answer:

Area = $9x^2$ square units

Step-by-step explanation:

<u>Key skills needed: Area of a parallelogram</u>

1) This shape is a parallelogram, and the way to find the area of a parallelogram is:

$A = bh$

A = area

b = base (the side at the bottom

h = the height ( the length that runs from one side to another and makes a right angle)

2) The base (b) would be 3x. The height (h) would also be 3x

3) This means A = $3x(3x)$

3 times 3 is 9 --> and x times x is $x^2$

Therefore the area is --> $9x^2$

<em>Hope you understood and have a nice day!! :D</em>

5 0

2 years ago

In right △ABC, the altitude CH to the hypotenuse AB intersects angle bisector AL in point D. Find the sides of △ABC if AD = 8 cm

tangare [24]

Answer:

$AC=8\sqrt{3}\ cm\\ \\AB=16\sqrt{3}\ cm\\ \\BC=24\ cm$

Step-by-step explanation:

Consider right triangle ADH ( it is right triangle, because CH is the altitude). In this triangle, the hypotenuse AD = 8 cm and the leg DH = 4 cm. If the leg is half of the hypotenuse, then the opposite to this leg angle is equal to 30°.

By the Pythagorean theorem,

$AD^2=AH^2+DH^2\\ \\8^2=AH^2+4^2\\ \\AH^2=64-16=48\\ \\AH=\sqrt{48}=4\sqrt{3}\ cm$

AL is angle A bisector, then angle A is 60°. Use the angle's bisector property:

$\dfrac{CA}{CD}=\dfrac{AH}{HD}\\ \\\dfrac{CA}{CD}=\dfrac{4\sqrt{3}}{4}=\sqrt{3}\Rightarrow CA=\sqrt{3}CD$

Consider right triangle CAH.By the Pythagorean theorem,

$CA^2=CH^2+AH^2\\ \\(\sqrt{3}CD)^2=(CD+4)^2+(4\sqrt{3})^2\\ \\3CD^2=CD^2+8CD+16+48\\ \\2CD^2-8CD-64=0\\ \\CD^2-4CD-32=0\\ \\D=(-4)^2-4\cdot 1\cdot (-32)=16+128=144\\ \\CD_{1,2}=\dfrac{-(-4)\pm\sqrt{144}}{2\cdot 1}=\dfrac{4\pm 12}{2}=-4,\ 8$