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OverLord2011 [107]

3 years ago

5

a student athlete runs 3 1/3 miles in 30 minutes a professional Runner can run one and one fourth times as far in 30 minutes how

far can the professional Runner run in 30 minutes

1 answer:

Readme [11.4K]3 years ago

7 0

Just multiply 3 1/3 x 1 1/4 which is 3 and 7/12 miles well together 3 , 7/12

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Math Correct answers only please!!

aksik [14]

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Problem 1</u>

Rewrite and add each vector:

$p=\langle40cos70^\circ,40sin70^\circ\rangle\\q=\langle50cos270^\circ,50sin270^\circ\rangle\\r=\langle60cos235^\circ,60sin235^\circ\rangle\\p+q+r=\langle40cos70^\circ+50cos270^\circ+60cos235^\circ,40sin70^\circ+50sin270^\circ+60sin235^\circ\rangle\\p+q+r\approx\langle-20.73,-61.56\rangle$

Find the magnitude of the resulting vector:

$||p+q+r||=\sqrt{x^2+y^2}\\||p+q+r||=\sqrt{(-20.73)^2+(-61.56)^2}\\||p+q+r||\approx64.96$

Therefore, the best answer is D) 64.959

<u>Problem 2</u>

Think of the vectors like this:

$t=\langle7cos240^\circ,7sin240^\circ\rangle\\u=\langle10cos30^\circ,10sin30^\circ\rangle\\v=\langle15cos310^\circ,15sin310^\circ\rangle$

By adding the vectors, we have:

$t+u+v=\langle7cos240^\circ+10cos30^\circ+15cos310^\circ,7sin240^\circ+10sin30^\circ+15sin310^\circ\rangle\\t+u+v\approx\langle14.8,-12.55\rangle$

Since the direction of a vector is $\theta=tan^{-1}(\frac{y}{x})$ , we have:

$\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{-12.55}{14.8})\\\theta\approx-40.3^\circ$

Don't forget to take into account that the resulting vector is in Quadrant IV since the horizontal component is positive and the vertical component is negative, so we will add 360° to our angle to get the result:

$\theta=360^\circ+(-40.3)^\circ\\\theta=319.7^\circ$

Therefore, the best answer is D) 320°

<u>Problem 3</u>

Again, rewrite the vectors and add them:

$u=\langle30cos70^\circ,30sin70^\circ\rangle\\v=\langle40cos220^\circ,40sin220^\circ\rangle\\u+v=\langle30cos70^\circ+40cos220^\circ,30sin70^\circ+40sin220^\circ\rangle\\u+v\approx\langle-20.38,2.48\rangle$

Using the direction formula:

$\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{2.48}{-20.38})\\\theta\approx-6.94^\circ$

As the vector is located in Quadrant II, we need to add 180° to our angle:

$\theta=180^\circ+(-6.94)^\circ\\\theta=173.06^\circ$

Therefore, the best answer is D) 173°

<u>Problem 4</u>

Using scalar multiplication:

$-3u=-3\langle35,-12\rangle=\langle-105,36\rangle$

Find the magnitude of the resulting vector:

$||-3u||=\sqrt{x^2+y^2}\\||-3u||=\sqrt{(-105)^2+(36)^2}\\||-3u||=111$

Find the direction of the resulting vector:

$\theta=tan^{-1}(\frac{y}{x})\\\theta=tan^{-1}(\frac{36}{-105})\\\theta\approx-18.92^\circ$

As the vector is located in Quadrant II, we need to add 180° to our angle:

$\theta=180^\circ+(-18.92)^\circ\\\theta=161.08^\circ$

Therefore, the best answer is C) 111; 161°

<u>Problem 5</u>

Using scalar multiplication:

$5v=5\langle25cos40^\circ,25sin40^\circ\rangle=\langle125cos40^\circ,125sin40^\circ\rangle$

Since the direction of the angle doesn't change in scalar multiplication, it only affects the magnitude, so the magnitude of the resulting vector would be $||5v||=125$ , making the correct answer C) 125; 200°

6 0

2 years ago

Amy and Laura recorded the average gas mileage of their vehicles each week for four weeks. What can we concluded about the relat

Assoli18 [71]

there is no table shown

4 0

3 years ago

Read 2 more answers

A chemist is mixing two solutions, solution A and solution B Solution A is 15% water and solution Bis 20% water. She already has

slavikrds [6]

Answer:

15 mL of the solution with 20% water will be needed.

Step-by-step explanation:

Use the inverse relationship

10 mL * (18-15)% = x mL * (20-18)%

x = 10 mL * (3/2) = 15 mL

4 0

3 years ago

Read 2 more answers

Multiply and divide. product or quotient

Mandarinka [93]

Multiply is product. Divide is quotient

7 0

2 years ago

Read 2 more answers

mark invests 8000 in an account that pays 12% interest and 2000 in one that pays 8%. if he leaves the money in the accounts for

frez [133]

mark must leave it for 5.5 months or 5 and half moths to gain 5600 in interest .

<u>Step-by-step explanation:</u>

Here we have , mark invests 8000 in an account that pays 12% interest and 2000 in one that pays 8%. if he leaves the money in the accounts for the same length of time, We need to find how long must he leave it to gain 5600 in interest . Let's find out:

Let mark invests 8000 in an account that pays 12% interest and 2000 in one that pays 8% for time x months , So total interest gain is 5600 i.e.

⇒ $\frac{8000(12)}{100}(x) +\frac{2000(8)}{100}(x) =5600$

⇒ $\frac{8000(12)+2000(8)}{100}(x) =5600$

⇒ $(80(12)+20(8))(x) =5600$

⇒ $(960+160)(x) =5600$

⇒ $(1020)(x) =5600$

⇒ $x =\frac{5600}{1020}$

⇒ $x =5.5$

Therefore , mark must leave it for 5.5 months or 5 and half moths to gain 5600 in interest .

6 0

3 years ago