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VARVARA [1.3K]
3 years ago
13

Use the law of cosines to find each missing side round to the nearest tenth

Mathematics
1 answer:
svet-max [94.6K]3 years ago
3 0

a^{2}  =  {b}^{2}  +  {c}^{2}  - 2bc \cos( \alpha)
a =  \sqrt{ {8}^{2} +  {14}^{2}  - 2(8)(14) \cos(67)  }
a = 13.13
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1 2 3 4 5 6 7 8 9 10
gayaneshka [121]

Answer:

The 8 should have remained positive.

Step-by-step explanation:

8-(-2)=8+2=10

Two negatives make a positive so 8-(-2) will turn into 8+2.

Laura was wrong because she thought the 8 turned to a negative but that makes no sense.

8 0
3 years ago
1. Solve for x. Show each step of the solution<br> 4.5(8 - x) + 36 = 102 - 2.5(3x + 24)
Vladimir [108]
36-4.5x+36=102-7.5x-60
72-4.5x=42-7.5x
72+3.5x=42
3.5x=-30
x=8.57
7 0
3 years ago
Mrs.Welch needs to buy some cat food . If each can costs $0.79 and she has a coupon for $2 off the entire purchase, how many can
allochka39001 [22]

Answer:

11 cans of cat food...........

Step-by-step explanation:

£0.79 x 11 = £8.69

£8.69 - £2 = 6.69

6 0
2 years ago
Read 2 more answers
13
il63 [147K]

The required value after simplification of the s = -16/3. None of these are correct.

Given that,
To simplify [\frac{x^{2/3}x^{-1/2}}{x\sqrt{x^3}\sqrt[3]{x}}]^2   and to find the value of s in x^s.

<h3>What is simplification?</h3>

The process in mathematics to operate and interpret the function to make the function simple or more understandable is called simplifying and the process is called simplification.

Simplification,

=[\frac{x^{2/3}x^{-1/2}}{x\sqrt{x^3}\sqrt[3]{x}}]^2\\= \frac{x^{4/3}x^{-1}}{x^2x^3*{x}^{2/3}}\\= \frac{x^{1/3}}{x^{17/3}}\\=x^{-16/3}
Comparing with x^S
s = -16/3


Thus, the required value of the s = -16/3. None of these are correct.

Learn more about simplification here: brainly.com/question/12501526

#SPJ1

4 0
1 year ago
True or false square root 54 is close to 7?
antoniya [11.8K]
No it's over : 7.3 , unless your prompts context is close to as in getting there or close in general?
3 0
3 years ago
Read 2 more answers
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