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vazorg [7]
3 years ago
6

In circle C, what is mArc F H? 31° 48° 112° 121°

Mathematics
2 answers:
faltersainse [42]3 years ago
3 0

Answer:

B. 48

Step-by-step explanation:

just took the test <3

Gemiola [76]3 years ago
3 0

Answer:

mArc FH = 48

Step-by-step explanation:

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2x+5(x+1) = -8;<br><br> Can someone help me solve this problem step by step please?
mariarad [96]

Answer:

-13/7

Step-by-step explanation:

2x+5(x+1)=-8

2x+5x+5=-8

7x+5=-8

7x=-8-5

7x=-13

7x/7=-13/7

x=-13/7

4 0
2 years ago
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1.15 + m = 10 Please give an answer
garik1379 [7]

Answer:

M = 8.85

Step-by-step explanation:

10 - 1.15 = 8.85

8 0
2 years ago
5y+6=-3y + 54 please help me solve this step by step asap !
Olenka [21]

Answer:

6=y

Step-by-step explanation:

5y + 6 = -3y +54

isolate y

+3y

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divide by 8 so y is alone w no coefficient other than 1

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3 years ago
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Geoffrey ran 13.5 miles in 1.5 hours. What was his speed in miles per hour?
Vlad1618 [11]

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0.0025 miles per second

Step-by-step explanation:

8 0
3 years ago
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What is the exact value of cos (67.5°)?
babymother [125]

first off, make sure you have a Unit Circle, if you don't do get one, you'll need it, you can find many online.

let's double up 67.5°, that way we can use the half-angle identity for the cosine of it, so hmmm twice 67.5 is simply 135°, keeping in mind that 135° is really 90° + 45°, and that whilst 135° is on the 2nd Quadrant and its cosine is negative 67.5° is on the 1st Quadrant where cosine is positive, so

cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\\\ cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}} \\\\[-0.35em] ~\dotfill\\\\ cos(135^o)\implies cos(90^o+45^o)\implies cos(90^o)cos(45^o)~~ - ~~sin(90^o)sin(45^o) \\\\\\ \left( 0 \right)\left( \cfrac{\sqrt{2}}{2} \right)~~ - ~~\left( 1\right)\left( \cfrac{\sqrt{2}}{2} \right)\implies -\cfrac{\sqrt{2}}{2} \\\\[-0.35em] ~\dotfill

cos(67.5^o)\implies cos\left( \frac{135^o}{2} \right)\implies \pm \sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}\implies \stackrel{I~Quadrant}{+\sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}} \\\\\\ \sqrt{\cfrac{ ~~ \frac{2-\sqrt{2}}{2} ~~ }{2}}\implies \sqrt{\cfrac{2-\sqrt{2}}{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{2}

8 0
2 years ago
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