Ask question

Ask question

All categories

3 years ago

13

Please Help!

1 answer:

Tamiku [17]3 years ago

4 0

Answer:

n

Step-by-step explanation:

the answer is -60, which is not an option

You might be interested in

I need help with this one

OlgaM077 [116]

Answer:

112.98 square units

Step-by-step explanation:

Let's work backwards:

Circumference = 2πr

(37.68)/2π = r

Area = π $r^{2}$

Area = π * $((37.68)/2pi)^{2}$

Area = 112.983 square units

3 0

3 years ago

Read 2 more answers

James invested $4,000 at 5% interest per year; how long will it take him to earn $200 in simple interest?

tester [92]

Answer:

It takes 1 year.

Step-by-step explanation:

You have to apply simple interest formula, I = (P×R×T)/100 where I represent interest amount, P is principle, R is interest rate and T is number of years :

$I = \frac{PRT}{100}$

$200 = \frac{4000 \times 5 \times T}{100}$

$20000 = 20000T$

$T = 1$

3 0

3 years ago

Read 2 more answers

Help please. I need the answer

grandymaker [24]

Answer:

y=-2/3x+6

Step-by-step explanation:

Graph it

6 0

3 years ago

Read 2 more answers

out of 30 students surveyed 17 have a dog based on these results predict how many of the 300 students in the school have a dog

Ira Lisetskai [31]

The whole school would have 170 students that have a dog

7 0

3 years ago

Read 2 more answers

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

NISA [10]

Answer:

$\theta = 108.29$

Step-by-step explanation:

Given

$u =$

$v =$

Required:

Calculate the angle between u and v

The angle $\theta$ is calculated as thus:

$cos\theta = \frac{u.v}{|u|.|v|}$

For a vector

$A =$

$A = a * b$

$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

$cos\theta = \frac{-16}{|u|.|v|}$

For a vector

$A =$

$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

<em></em> $\theta = 108.29$ <em> (approximated)</em>

4 0

2 years ago