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Liula [17]
3 years ago
10

Find a. Roud to the nearest tenth: 2cm 15* a 105* c

Mathematics
1 answer:
MariettaO [177]3 years ago
6 0

Answer:

So since we use law of sines. We can get 2 fractions equal to eachother but first we need to solve for missing angle. That would be 180-105-15 = 60

So now we can get a fraction. So that means that sin(105)/2 = sin(60)/a

Then we do recipricol. 2/sin(105) = a/sin(60)

multiply by sin(60) both sides.

2sin(60)/sin(105) = a

round and solve and we get a is equal to

1.7931 or 1.8

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So scientific notation is making the number manageable.

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(a number >1 and <10) times 10^x

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the answer is 6.
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8 0
3 years ago
In the diagram shown, lines m and n are
bonufazy [111]

Answer:

Step-by-step explanation:

If two line 'm' and 'n' are parallel and a transversal line 't' i intersecting these lines at two distinct points,

Sum of interior consecutive angles should be 180°.

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Therefore, 114° + 64° = 180°

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3 0
3 years ago
Help! I will Mark Brainliest!
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Answer:

Your answer is B. I had some help from my mom.

Step-by-step explanation:

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5 0
2 years ago
The model below can be used to find the quotient of one over two divided by one over six. What is the quotient?
VLD [36.1K]

Answer:

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Step-by-step explanation:

3 0
3 years ago
Mica bought a new smart phone for $258.13. That was the price after his 18.25% discount. What was the original price of the phon
julia-pushkina [17]

The original price of phone is $ 315.76

<em><u>Solution:</u></em>

Given that Mica bought a new smart phone for $ 258.13.

That was the price after his 18.25% discount

<em><u>To find: original price of phone</u></em>

From given question,

price after discount = $ 258.13

discount = 18.25 %

Let "x" be the original price of phone

price after discount = original price - discount

258.13 = x - 18.25 \% \text{ of } x

258.13 = x - \frac{18.25}{100}x

258.13 = x(1 - \frac{18.25}{100})

258.13 = x(1 - 0.1825)\\\\258.13 = x( 0.8175)\\\\x = \frac{258.13}{0.8175}\\\\x = 315.76

Thus original price of phone is $ 315.76

4 0
4 years ago
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