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4 years ago

(7)(8)(9) Geometry Check Ans 7) s = 15 8) y = 64 x = 99 9) 96

Mathematics

1 answer:

Katyanochek1 [597]4 years ago

6 0

Problem 7: Correct
Problem 8: Correct
Problem 9: Correct

The steps are below if you are curious

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Problem 7

S = 180*(n-2)
2340 = 180*(n-2)
2340/180 = n-2
13 = n-2
n-2 = 13
n = 13+2
n = 15

I'm using n in place of lowercase s, but the idea is the same. If anything, it is better to use n for the number of sides since S already stands for the sum of the interior angles. I'm not sure why your teacher decided to swap things like that.

===========================================================================================

Problem 8

First find y
y+116 = 180
y+116-116 = 180-116
y = 64

which is then used to find x. The quadrilateral angles add up to 180*(n-2) = 180*(4-2) = 360 degrees
Add up the 4 angles, set the sum equal to 360, solve for x

x+y+125+72 = 360
x+64+125+72 = 360 ... substitution (plug in y = 64)
x+261 = 360
x+261-261 = 360-261
x = 99

===========================================================================================

Problem 9

With any polygon, the sum of the exterior angles is always 360 degrees

The first two exterior angles add to 264. The missing exterior angle is x
x+264 = 360
x+264-264 = 360-264
x = 96

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are you sure you wrote the problem here correctly ?

because the distance will be 40km after less than half an hour just by the first car driving. way before the second car even starts.

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3 years ago

How do I complete these questions a bit confused . (extra credit work )

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Answer:

(1)

$a = \frac{3\sqrt 3}{2}$

$b = \frac{3}{2}$

(2)

$a = \sqrt 6$

$b = \sqrt 2$

Step-by-step explanation:

Solving (1):

Considering

$\theta = 60^o$

We have:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

This gives:

$\sin(60^o) = \frac{a}{3}$

Solve for a

$a = 3 * \sin(60^o)$

$\sin(60^o) = \frac{\sqrt 3}{2}$

So:

$a = 3 * \frac{\sqrt 3}{2}$

$a = \frac{3\sqrt 3}{2}$

To solve for b, we make use of Pythagoras theorem

$3^2 = a^2 + b^2$

This gives

$3^2 = (\frac{3\sqrt 3}{2})^2 + b^2$

$9 = \frac{9*3}{4} + b^2$

$9 = \frac{27}{4} + b^2$

Collect like terms

$b^2 = 9 - \frac{27}{4}$

Take LCM and solve

$b^2 = \frac{36 - 27}{4}$

$b^2 = \frac{9}{4}$

Take square roots

$b = \frac{3}{2}$

Solving (2):

Considering

$\theta = 60^o$

We have:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

This gives:

$\sin(60^o) = \frac{a}{2\sqrt 2}$

Solve for a

$a = 2\sqrt 2 * \sin(60^o)$

$\sin(60^o) = \frac{\sqrt 3}{2}$

So:

$a = 2\sqrt 2 * \frac{\sqrt 3}{2}$

$a = \sqrt 2 * \sqrt 3$

$a = \sqrt 6$

To solve for b, we make use of Pythagoras theorem

$(2\sqrt 2)^2 = a^2 + b^2$

This gives

$(2\sqrt 2)^2 = (\sqrt 6)^2 + b^2$

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Collect like terms

$b^2 = 8 - 6$

$b^2 = 2$

Take square roots

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