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3 years ago

The least common multiple of 4 and 7

Mathematics

1 answer:

shusha [124]3 years ago

5 0

The answer is 28. My answer is correct. 7, 14& 21 can't be the LCM son the answer should be 28.

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Barney is the owner and insured of a $200,000 life insurance policy. Upon learning that he has a terminal illness, he sells the

ludmilkaskok [199]

Answer:

Viatical settlement

Step-by-step explanation:

Here Barney sold his life insurance to 3rd party for less net death benefit value.

He entered into the arrangement called - viatical settlement

A viatical settlement is the sale of a life insurance policy to a third party. The owner of the life insurance policy sells the policy for cash benefit, thus making the buyer the new owner of the policy who will get death benefits.

4 0

3 years ago

Plz help me with this I don't have my protractor with me plz help and do this

My name is Ann [436]

An acute angle is an angle less than 90 degrees. (smaller angle)
A right angle is exactly 90 degrees. (right angle)
An obtuse angle is over 90 degrees. (larger angle)
A reflex angle is any angle of 180 degrees. (largest angle)

You wouldn't *really* need a protractor to complete your homework; you could guesstimate.

I hope this helps!

8 0

4 years ago

PLEASE HELP

riadik2000 [5.3K]

Answer:

1 and -3

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the answer?<br> -1.2x=12

rjkz [21]

775757874875875345634634563456

3 0

3 years ago

Find the absolute maximum and minimum values of f on the set D. f(x,y)=2x^3+y^4, D={(x,y) | x^2+y^2<=1}.

castortr0y [4]

Answer:

absolute maximum is f(1, 0) = 2 and the absolute minimum is f(−1, 0) = −2.

Step-by-step explanation:

We compute,

$f_x = 6x^2, f_y=4y^3$

Hence, $f_x = f_y = 0$ if and only if (x,y) = (0,0)

This is unique critical point of D. The boundary equation is given by

$x^2+y^2=1$

Hence, the top half of the boundary is,

$$ T = \{ x, \sqrt{1-x^2} : -1 \leq x \leq 1\}$

On T we have, $f(x, \sqrt{1-x^2} = 2x^3 +(1-x^2)^2 = x^4 +2x^3-2x^2+1 \text{ for}\ -1 \leq x \leq 1$

We compute

$\frac{d}{dx}(f(x, \sqrt{1-x^2}))= 4x^3+6x^2-4x = 2x(2x^2+3x-2)=2x(2x-1)(x+2)=0$

0 if and only if x=0, x= 1/2 or x = -2.

We disregard $x = -2 \notin [-1,1]$

Hence, the critical points on T are (0,1) and $(\frac{1}{2}, \sqrt{1-(\frac{1}{2})^2}=\frac{\sqrt3}{2})$

On the bottom half, B, we have

$f(x, \sqrt{1-x^2})= f(x,-\sqrt{1-x^2})$

Therefore, the critical points on B are (0,-1) and $$( 1/2, -\sqrt3/2)$

It remains to evaluate f(x, y) at the points $(0,0), (0 \pm1), (1/2, \pm \sqrt3/2) \text{ and}\ (\pm1, 0)$ .

We should consider latter two points, $(\pm1,0)$ , since they are the boundary points for the T and also B. We compute $f(0,0)=0, \ \f(0 \pm1)=1, \ \ f(0, \pm \sqrt3/2)=9/16, \ \ f(1,0 )= 2 \text{ and}\ \ f(-1,0)= -2$

We conclude that the absolute maximum = f(1, 0) = 2

And the absolute minimum = f(−1, 0) = −2.

6 0

3 years ago