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olchik [2.2K]
2 years ago
9

8-4 trigonometry (part b) hw (2nd part)

Mathematics
1 answer:
marishachu [46]2 years ago
7 0

Answer:

82+25=107 yan po ang answer

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Artemon [7]
C is the correct answer
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3 years ago
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What is the solution of the system?<br> y=x^2-4x+5<br> Y=-2+8
Nataly_w [17]
Yea something is not right.if y=6 then you cant solve the system because you end up with x^2-4x-1
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3 years ago
Need help with these
NNADVOKAT [17]
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7 0
3 years ago
the slope of a line passing through h (-2, 5) is -3/4. Which ordered pair represents a point on this line? ​
rosijanka [135]

Answer:

A

Step-by-step explanation:

Calculate the slope of the given points with the point (- 2,  5 )

If the slope is - \frac{3}{4} then the point is on the line

Calculate slope m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (6, - 1)

m = \frac{-1-5}{6+2} = \frac{-6}{8} = - \frac{3}{4} ← point (6, - 1) is on the line

Repeat with (x₁, y₁ ) = (- 2, 5 ) and (x₂, y₂ ) = (2, 8)

m = \frac{8-5}{2+2} = \frac{3}{4} ← point (2, 8) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (- 5, 1)

m = \frac{1-5}{-5+2} = \frac{-4}{-3} = \frac{4}{3} ← point (- 5, 1) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (1, 1)

m = \frac{1-5}{1+2} = - \frac{4}{3} ← point (1, 1) is not on the line

Thus the point on the line is (6, - 1 ) → A

4 0
2 years ago
Analysis of an accident scene indicates a car was traveling at a velocity of 69.5 mph (31.1 m/s) along the positive x-axis at th
STatiana [176]

We can find the acceleration via

{v_f}^2-{v_i}^2=2a\Delta x

We have

\left(5.20\dfrac{\rm m}{\rm s}\right)^2-\left(31.1\dfrac{\rm m}{\rm s}\right)^2=2a(115\,\mathrm m)

\implies\boxed{a=-4.09\dfrac{\rm m}{\mathrm s^2}}

Then by definition of average acceleration,

a_{\rm ave}=\dfrac{v_f-v_i}t

so that

-4.09\dfrac{\rm m}{\mathrm s^2}=\dfrac{5.20\frac{\rm m}{\rm s}-31.1\frac{\rm m}{\rm s}}t

\implies\boxed{t=6.33\,\mathrm s}

We alternatively could have found the time without knowing the acceleration. Since acceleration is constant, the average velocity is

v_{\rm ave}=\dfrac{x_f-x_i}t=\dfrac{v_f+v_i}2

Then

\dfrac{115\,\rm m}t=\dfrac{5.20\frac{\rm m}{\rm s}+31.1\frac{\rm m}{\rm s}}2

\implies\boxed{t=6.33\,\mathrm s}

7 0
3 years ago
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