All categories

3 years ago

Solve for c: 6/15=2/c

Mathematics

1 answer:

Solnce55 [7]3 years ago

7 0

6/15 = 2/c

c = 2/6 * 15

c = 5

You might be interested in

Simplify remove all the perfect squares from inside the square root 28

expeople1 [14]

Answer:

$2\sqrt{7}$

4 0

3 years ago

If m∠NOP = 24° and m∠NOQ = 110°, what is m∠POQ?

monitta

Correct answer for the above question is - option B. 86°

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

Given:

∠NOP = 24°

∠NOQ = 110°

∠NOP and ∠POQ are adjacent angles

To Find:

∠POQ = ?

Solution:

Hereby, we can say that ∠POQ lies between line OQ and ON as given (∠NOP and ∠POQ are adjacent angles )

∠NOQ - ∠NOP = ∠POQ

∠POQ = 110° - 24°

Angle POQ is 86°

Thus we can conclude option B as correct answer.

3 0

2 years ago

Can anyone help me please

Xelga [282]

popopoppopopoopopopp[poopopopopopop

3 0

2 years ago

What is 16,107,320 written in word form

Colt1911 [192]

Sixteen million one hundred seven thousand three hundred and twenty

8 0

2 years ago

To find the extreme values of a function f(x,y) on a curve xequalsx(t), yequalsy(t), treat f as a function of the single vari

pychu [463]

Answer:

Absolute maximum is 2

Absolute minimum at -2

Step-by-step explanation:

The given parametric functions are:

$x=2\cos t,y=2\sin t$

By the chain rule:

$f'(t)=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{df}{dt} =\frac{2 \cos t}{-2\sin t} =-\cot(t)$

At fixed points, $f'(t)=0$

$\implies -\cot (t)=0$

This gives $t=\frac{\pi}{2} ,\frac{3\pi}{2}$ on $0\le t\le 2\pi$

This implies that the extreme points are $(2\cos \frac{\pi}{2}, 2\sin \frac{\pi}{2})=(0,2)$ and $(2\cos \frac{3\pi}{2}, 2\sin \frac{3\pi}{2})=(0,-2)$

By eliminating the parameter, we have $x^2+y^2=4$

This is a circle with radius 2, centered at the origin.

Hence (0,2) is an absolute maximum ,at $t=\frac{\pi}{2}$ and (0,-2) is an absolute minimum at $t=\frac{3\pi}{2}$

7 0

2 years ago