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padilas [110]
3 years ago
13

Jessica left her running shoes at school yesterday. Today she walked 4 miles to school to get her shoes, she ran home along the

same route, and the total time for both trips was 2 hours. Jessica walked and ran at constant speeds, and she ran 3 miles per hour faster than she walked.
What was Jessica's walking speed in miles per hour?
Mathematics
2 answers:
Vikentia [17]3 years ago
7 0

Jessica left her running shoes at school yesterday. Today she walked 4 miles to school to get her shoes, she ran home along the same route, and the total time for both trips was 2 hours. Jessica walked and ran at constant speeds, and she ran 3 miles per hour faster than she walked.

What was Jessica's walking speed in miles per hour?

ehidna [41]3 years ago
3 0

Answer:

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\small\textrm\green{-------------------------------------------------------------------}\small\textrm\blue{-------------------------------------------------------------------}\small\textrm\green{-------------------------------------------------------------------}\small\textrm\blue{-------------------------------------------------------------------}\small\textrm\green{-------------------------------------------------------------------}\small\textrm\blue{-------------------------------------------------------------------}

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(-4,0) (3,-5) I don't know if they mean two ordered pairs so I'm pretty sure this is right
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Which correspondence accurately describes the two congruent triangles shown here?
Crazy boy [7]
Your answer is correct
4 0
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Write a linear equation to represent the line shown on the graph
sveticcg [70]

Answer:

y=-3x+4

Step-by-step explanation:

it's negative bc the slope is going from left to right. you just have to do rise over run to find the slope.

7 0
3 years ago
Which statements about the hyperbola are true? Check all that apply. A. There is a focus at (0,−10). B. There is a focus at (0,
hichkok12 [17]

Answer:

A. There is a focus at (0,−10).

Step-by-step explanation:

Assume the hyperbola is like the one below.

The hyperbola is vertical and centred on the y-axis, so its general equation is

\dfrac{y^{2}}{a^{2}} - \dfrac{x^{2}}{b^{2}} = 1

The vertices of your parabola are (0,±8) so a = 8.

The covertices are (±6,0), so b = 6.

Calculate c

\begin{array}{rcl}a^{2} + b^{2} & = & c^{2}\\8^{2} + 6^{2} & = & c^{2}\\64 + 36 & = & c^{2}\\100 & = & c^{2}\\c & = & \mathbf{10}\\\end{array}

A. Foci

The foci are at (0, ±c) = (0, ±10)

TRUE. There is a focus at (0,-10).

B. Foci

The foci are at (0,±10).

False. There is no focus at (0,12)

C. and D. Asymptotes

The equations for the asymptotes are

y = \pm\dfrac{a}{b}x = \pm\dfrac{8}{6}x = \pm\dfrac{4}{3}x

So, y = ±x are not asymptotes.

False.

E. and F. Directrices

The equations of the directrices are  

y = ±a²/c = ±64/10 = ±6.4

y = 6.4 is a directrix.

E is false. x = cannot be a directrix

F is uncertain. Your equation for the directrix is incomplete.

3 0
4 years ago
Consider a triangle ABC like the one below. Suppose that a = 31, b = 23, and c = 20. (The figure is not drawn to scale.) Solve t
krok68 [10]

The solution to the triangle is A = 92.0, B = 47.9 and C = 40.1

<h3>How to solve the triangle?</h3>

The figure is not given;

However, the question can still be solved without it

The given parameters are:

a = 31, b = 23, and c = 20

Calculate angle A using the following law of cosine

a² = b² + c² - 2bc * cos(A)

So, we have:

31² = 23² + 20² - 2 * 23 * 20 * cos(A)

Evaluate

961 = 929 - 920 * cos(A)

Subtract 929 from both sides

32 =- 920 * cos(A)

Divide both sides by -920

cos(A) = -0.0348

Take the arc cos of both sides

A = 92.0

Calculate angle B using the following law of sine

a/sin(A) = b/sin(B)

So, we have:

31/sin(92) = 23/sin(B)

This gives

31.0189 = 23/sin(B)

Rewrite as:

sin(B) =23/31.0189

Evaluate

sin(B) =0.7415

Take arc sin of both sides

B = 47.9

Calculate angle C using:

C = 180 - 92.0 - 47.9

Evaluate

C = 40.1

Hence, the solution to the triangle is A = 92.0, B = 47.9 and C = 40.1

Read more about triangles at:

brainly.com/question/2217700

#SPJ1

6 0
2 years ago
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