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2 years ago

15

Hector runs laps around the track. The table shows how long it takes him to run different numbers of laps. How long would it tak

e Hector to run 7 laps?
A.35 minutes

B.25 minutes

C. 15 minutes

D. 45 minutes

2 answers:

lorasvet [3.4K]2 years ago

6 0

You were supposed to put a picture of the graph

Alex777 [14]2 years ago

4 0

I don’t see a graph so i can’t answer

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Complete the following steps to find the LCD and write the sum of the numerators for the given problem: es002-1. Jpg Factor each

lesya [120]

The value of x from the given equation is -4/3

<h3>Factorization</h3>

Given the quadratic equation x^2 – 3x – 4 = x^2 – 6x + 8

Collect the like terms

x^2 – 3x – 4 - x^2 + 6x - 8 = 0

-3x + 6x - 4 + 8 = 0

3x + 4 = 0

Subtrcat 4 from both sides

3x + 4 - 4 = 0 - 4

3x = -4

Divide both sodes by 3

3x/3 = -4/3

x = -4/3

Hence the value of x from the given equation is -4/3

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2 years ago

Help 9-11 thank you !

krok68 [10]

Answer:

9 - 11 = -2

Step-by-step explanation:

Because 9 is being subtracted by a number greater than itself, the answer has to be a negative and because:

9 + 2 = 11

9 - 11 = (-2)

Your Welcome :)

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2 years ago

-----40 POINTS--------

andrey2020 [161]

Proof

Provide the missing reasons for the proof of part of the triangle midsegment theorem.

Given: K is the midpoint of MJ.

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The complete answer for the missing reasons is attached below in the diagram.

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3 years ago

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7 1/2 pints = _____ quarts

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3 years ago

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HELP! Find the value of sin 0 if tan 0 = 4; 180 < 0< 270

BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

$\large \boxed{sin \theta = tan \theta cos \theta} \\ \large \boxed{tan^{2} \theta + 1 = {sec}^{2} \theta}$

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

$\large{ {4}^{2} + 1 = {sec}^{2} \theta } \\ \large{16 + 1 = {sec}^{2} \theta } \\ \large{ {sec}^{2} \theta = 17}$

As we know, sec²θ = 1/cos²θ.

$\large \boxed{sec \theta = \frac{1}{cos \theta} } \\ \large \boxed{ {sec}^{2} \theta = \frac{1}{ {cos}^{2} \theta} }$

And thus,

$\large{ {cos}^{2} \theta = \frac{1}{17}} \\ \large{cos \theta = \frac{ \sqrt{1} }{ \sqrt{17} } } \\ \large{cos \theta = \frac{1}{ \sqrt{17} } \longrightarrow \frac{ \sqrt{17} }{17} }$

Since the given domain is 180° < θ < 360°. Thus, the cosθ < 0.

$\large{cos \theta = \cancel\frac{ \sqrt{17} }{17} \longrightarrow cos \theta = - \frac{ \sqrt{17} }{17}}$

Then use the Identity of sinθ = tanθcosθ to find the sinθ.

$\large{sin \theta = 4 \times ( - \frac{ \sqrt{17} }{17}) } \\ \large{sin \theta = - \frac{4 \sqrt{17} }{17} }$

Answer

sinθ = -4sqrt(17)/17 or A choice.

4 0

3 years ago