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mixas84 [53]
3 years ago
13

Please Help ASAP! Would be greatly appreciated.

Mathematics
2 answers:
Ymorist [56]3 years ago
8 0
Your answer would be G. You can do away with F and H immediately, because these line graphs show you no important information, and graph J shows 24/3, rather than 24/8, so it is incorrect also.
natima [27]3 years ago
6 0
Hi,
The answer G represents -24/8 the best because it is dividing it into eight parts.
<span>Hope this helps you.</span>
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Solve.<br> Subtract 6x2 – X – 7 from the product of x2 + x + 5 and x - 6.
tresset_1 [31]

Answer:

x ^ 3 − 11 x ^ 2 + X - x − 23

Step-by-step explanation:

you have to Simplify the expression. Hope that helped!! :)

PLEASE MARK BRAINLIEST!!

8 0
3 years ago
What is the range of this function
Marina86 [1]

The range of a function is the output values ( Y values)

These would be the numbers the arrows are pointing at.

-8, -3 , 5 and 7

The answer is C.

3 0
3 years ago
The front tent is 8 feet and 5 feet tall what is the part of the tent
KIM [24]
Well if u think about it 8*5= 40ft ²
6 0
3 years ago
Finding Hypotenuse Lengths. Find the length of the hypotenuse.
ioda

Answer:

11) \frac{3\sqrt{5}}{2}

12) 4\sqrt{5}

13) 22

14) 18

15) 2

16) 24\sqrt{3}

Step-by-step explanation:

For these problems I used the pythagorean theorem: a^{2}+ b^{2}= c^{2} and SOHCAHTOA

Sin = \frac{opposite}{hypotenuse}

Cos = \frac{adjacent}{hypotenuse}

Tan = \frac{opposite}{hypotenuse}

11)

First find the length of the bottom side by using Cos

cos(60)=\frac{x}{3}

3(cos(60))=x

1.5=x

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

3^{2}+ 1.5^{2}= c^{2}

9+2.25=c^{2}

\sqrt{11.25}=\sqrt{c^{2}}

\frac{3\sqrt{5}}{2}

12)

Find the length of the bottom side using Cos

cos(60)=\frac{x}{8}

8(cos(60))=x

4=x

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

8^{2}+ 4^{2}= c^{2}

64+16=c^{2}

\sqrt{80} =\sqrt{c^{2} }

4\sqrt{5}

13)

Find the length of the other side by using Tan

tan(30)=\frac{x}{11\sqrt{3} }

11\sqrt{3}* (tan(30)=x

11=x

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

(11\sqrt{3}) ^{2}+ 11^{2}= c^{2}

363+121=c^{2}

\sqrt{484} =\sqrt{c^{2}}

22

14)

(This is probably an easier way to do these problems)

Find the hypotenuse by using Cos (\frac{adjacent}{hypotenuse})

cos(60)=\frac{9}{x}

cos(60)x=9

x=\frac{9}{cos(60)}

x=18

15)

Find the hypotenuse using Sin (\frac{opposite}{hypotenuse})

sin(30)=\frac{1}{x}

sin(30)x=1\\x=\frac{1}{sin(x)}

x=2

16)

Find the hypotenuse using Cos (\frac{adjacent}{hypotenuse})

cos(60)=\frac{12\sqrt{3} }{x}

cos(60)x=12\sqrt{3}

x=\frac{12\sqrt{3} }{cos(60)}

x=24\sqrt{3}

5 0
4 years ago
I need help, please, I really need a help, please.
Vanyuwa [196]
The Answer Would Be CF
6 0
4 years ago
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