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Pachacha [2.7K]
3 years ago
11

Need help here please :)

Mathematics
1 answer:
nalin [4]3 years ago
4 0
1. A.Use law of cosines. cosA=(b^2+c^2-a^2)/(2bc) because A is the included angle between b and c. Plug in A=50 degrees, b=13, c=6. cos50=(13^2+6^2-a^2)/(2*13*6), a^2=104.7,a=10.2 approximately, so choose A.

2. A.Use law of cosines again. cosC=(a^2+b^2-c^2)/(2ab). C=95 degrees, a=12, b=22. Plug in, cos95=(12^2+22^2-c^2)/(2*12*22), solve the equation, c^2=674, c=26 approximately. The use law of sines to solve for angle A (also works for B), a/sinA=c/sinC, 12/sinA=26/sin95, sinA=0.46, A=arcsin(0.46)=27.6. Choose A.

3. Answer is A. Area=1/2bc*sinA, since A is the included angle between b and c. Plug in b=30, c=14, A=50 degrees, area=1/2*14*30*sin50=160. 87, so the answer is A.

4. D. As long as the sum of any two sides of the triangle is bigger than the third, the triangle exists. 240+121>263, 240+263>121, 263+121>240, so it exists. To use Heron's formula, first find the semiperimeter, (240+263+121)/2=312. A=\sqrt(312*(312-240)*(312-263)*(312-121))=14499.7 approximately, so choose D.

5. 300. The included angle between the two paths is C=49.17+90=139.17 degrees. The lengths of the two paths are a=150, b=170. c is the distance we want. Use the law of cosines, cosC=(a^2+b^2-c^2)/(2ab). Plug in, c^2=89989, c=300 approximately.
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  4877 fish

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Set each piece = to 56, assuming that the 2 variables for which you are not solving are each equal to 0.

7x -2(0)-14(0) = 56
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X = ?
(__, 0, 0)

7(0) - 2y - 14 (0) = 56
-2y = 56 (divide each side by -2)
Y= ?
(0, __ , 0)

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Anni [7]

Answer:

y=-3x+7

Step-by-step explanation:

y=mx+b is the formula you are going to end up once completed.

This, by given two points:

(4,-5) (3,-2)

\frac{-2-5}{3-4} = Point-Slope Intercept Form

\frac{3}{-1} = After subtracting both from above

-3 = is your slope, decreasing.

Substitute -3 for y=mx+b

y=-3x+b

Use one of the given points (any) to find what b equals.

(4,-5)

x    y

-5=-3(4)+b

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Final Equation:

y=-3x+7

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2 years ago
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