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2 years ago

8

The system of equations

1 answer:

Alex_Xolod [135]2 years ago

7 0

Answer:

no solution

Step-by-step explanation:

y = x+1 and y = x + 2 don't touch because they are parallel

if you don't know what parallel means, just search it up, but i can explain it to you:

parallel means same slope or same steepness of line

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2 years ago

Segment EF is the midsegment of trapezoid ABCD. Find the length of segment EF if segment AB is 20 and segment CD is 12.

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Midsegment = 1/2(base1 + base2)

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Answer

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2 years ago

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3 years ago

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