All categories

3 years ago

Evaluate your expression for m =12

1 answer:

3 0

There's no way for me to do that, because my expression
is totally blank, and doesn't involve ' m ' in any way.

But if you'll come back and give us your expression, I'll
evaluate it for m=12, and I'll also show you how.

You might be interested in

A dog weighs two pounds less than three times the weight of a cat. The dog also weights twenty-two more pounds than the cat. Wri

Solnce55 [7]

Answer:

the cat weighs 21 pounds.

the dog weighs 43 pounds

Step-by-step explanation:

dog = 3x-20 = x + 22

cat = x

3x-20 = x+22

3x = x+42 (add 20 to both sides)

2x = 42 (subtract x from both sides)

x = 21

4 0

3 years ago

Find an irrational number that is between 5.2 and 5.5. Explain why it is irrational. Include the decimal approximation of the ir

alexgriva [62]

Answer:

An apple, potato, and onion all taste the same if you eat them with your nose plugged

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

3) Let h(x)= |2x|-4. Find h(x) when x = -7. a) -10 b) 10 c) -13 d) 13

svlad2 [7]

Answer:

$h(x) = 10$

Step-by-step explanation:

Given

$h(x) = |2x| - 4$

$x = -7$

Find h(x)

Substitute -7 for x

$h(x) = |2*-7| - 4$

$h(x) = |-14| - 4$

Absolute values return positive. So:

$h(x) = 14 - 4$

$h(x) = 10$

8 0

2 years ago

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

7 0

3 years ago

Read 2 more answers

How many x intercepts appear on the graph of this polynomial function? f(x)=x^4-x^3+x^2-x

elena55 [62]

I believe that the answer is 3 intercepts

7 0

3 years ago