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2 years ago

9

Mr. Mole's burrow lies 5 55 meters below the ground. He started digging his way deeper into the ground, descending 3 33 meters e

ach minute.

1 answer:

Alexeev081 [22]2 years ago

7 0

Answer:

184,815

Step-by-step explanation:

so my answer my not be right so i would say double check but anyway just multiply 3333 and 555 and this should be your answer.

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I need help with questions #7 and #8 plz

katen-ka-za [31]

Answer:

7. A = 40.8 deg; B = 60.6 deg; C = 78.6 deg

8. A = 20.7 deg; B = 127.2 deg; C = 32.1 deg

Step-by-step explanation:

Law of Cosines

$c^2 = a^2 + b^2 - 2ab \cos C$

You know the lengths of the sides, so you know a, b, and c. You can use the law of cosines to find C, the measure of angle C.

Then you can use the law of cosines again for each of the other angles. An easier way to solve for angles A and B is, after solving for C with the law of cosines, solve for either A or B with the law of sines and solve for the last angle by the fact that the sum of the measures of the angles of a triangle is 180 deg.

7.

We use the law of cosines to find C.

$18^2 = 12^2 + 16^2 - 2(12)(16) \cos C$

$324 = 144 + 256 - 384 \cos C$

$-384 \cos C = -76$

$\cos C = 0.2$

$C = \cos^{-1} 0.2$

$C = 78.6^\circ$

Now we use the law of sines to find angle A.

Law of Sines

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

We know c and C. We can solve for a.

$\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$

$\dfrac{12}{\sin A} = \dfrac{18}{\sin 78.6^\circ}$

Cross multiply.

$18 \sin A = 12 \sin 78.6^\circ$

$\sin A = \dfrac{12 \sin 78.6^\circ}{18}$

$\sin A = 0.6535$

$A = \sin^{-1} 0.6535$

$A = 40.8^\circ$

To find B, we use

m<A + m<B + m<C = 180

40.8 + m<B + 78.6 = 180

m<B = 60.6 deg

8.

I'll use the law of cosines 3 times here to solve for all the angles.

Law of Cosines

$a^2 = b^2 + c^2 - 2bc \cos A$

$b^2 = a^2 + c^2 - 2ac \cos B$

$c^2 = a^2 + b^2 - 2ab \cos C$

Find angle A:

$a^2 = b^2 + c^2 - 2bc \cos A$

$8^2 = 18^2 + 12^2 - 2(18)(12) \cos A$

$64 = 468 - 432 \cos A$

$\cos A = 0.9352$

$A = 20.7^\circ$

Find angle B:

$b^2 = a^2 + c^2 - 2ac \cos B$

$18^2 = 8^2 + 12^2 - 2(8)(12) \cos B$

$324 = 208 - 192 \cos A$

$\cos B = -0.6042$

$B = 127.2^\circ$

Find angle C:

$c^2 = a^2 + b^2 - 2ab \cos C$

$12^2 = 8^2 + 18^2 - 2(8)(18) \cos B$

$144 = 388 - 288 \cos A$

$\cos C = 0.8472$

$C = 32.1^\circ$

8 0

2 years ago

PLS HELP! ASAP!<br><br> Find the value of x in the triangle below

Aloiza [94]

Answer:

the answer is 56

Step-by-step explanation:

because it is an equilateral triangle the other side is 62 so you add 62+62=124 180-124=56

7 0

2 years ago

Please answer all of the questions and if u are correct I will give u brainliest ;) xoxo <3

garik1379 [7]

Answer:

the answer for the 2nd question is 1/4

Step-by-step explanation:

i took the test

3 0

1 year ago

Can someone help me understand how to create 3 proofs

Katarina [22]

Step-by-step explanation:

If I'm understanding what a "proof" means, some answers could be:

2 + 3 = 5

4 + 15 = 19

8 + 21 = 29

Hope this helps you! Have a good night!

7 0

2 years ago

Given that P = (2, 9) and Q = (4, 14), find the component form and magnitude of vector PQ.

vredina [299]

The magnitude of the vector is
$\sqrt{2 {}^{2} + 5 {}^{2} } = \sqrt{29}$
the component form is
<2,5>

8 0

3 years ago

Read 2 more answers