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4 years ago

Please help me with this

Mathematics

1 answer:

kondaur [170]4 years ago

4 0

For number 1a. the answer is 7 to 26 or
7:26 or
$\frac{7}{26}$
For number 2a. the answer is 8 to 28 or
8:28 or
$\frac{8}{28} or \frac{2}{7}$
For number 3a. the answer is 12 to 13 or
12:13 or
$\frac{12}{13}$
For number 4a. the answer is 19 to 12 or
19:12 or
$\frac{19}{12}$
For number 5a. the answer is 29 to 10 or
29:10 or
$\frac{29}{10}$
For number 6a. the answer is 11 to 28 or
11:28 or
$\frac{11}{28}$

I just did your homework

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aksik [14]

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

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

Problem 2

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°

Problem 3

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°

Problem 4

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°

Problem 5

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

H - 20/j = k solve for j

mina [271]

Answer: j = 20/k-h

Step-by-step explanation: Multiply to remove your variable from your denominator.

I hope this helps you out.

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3 years ago

I WILL GIVE A BRAIN LIST TO THE FIRST PERSON WHO ANSWERS!!!!!!!

aksik [14]

Answer:

its c

Step-by-step explanation:

Our denominators are 40 and 4. What we need to do is find the lowest common denominator of the two numbers. This is the smallest number that can be divided by both 40 and 4. In this case, the lowest common denominator is 40.

If we multiply the first denominator (40) by 1 we will get 40. If we multiply the second denominator (4) by 10 we will also get 40. We also need to multiply the numerators above the line by the same amounts so that the fraction values are correct:

33 x 1

40 x 1

3 x 10

4 x 10

This is what 33/40 and 3/4 looks like with the same denominator:

33

30

Now that these fractions have been converted to have the same denominator, we can clearly see by looking at the numerators that 33 is greater than 30 which also means that 33/40 is greater than 3/4.

6 0

3 years ago

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