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Free points what is 2+2

olganol [36]

Answer:

4

Step-by-step explanation:

thanks so much 2+2 equals 4 because that what it equals theres bo ther way

5 0

2 years ago

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What is the first step in solving quadratic equations by finding square roots? a. give the positive and negative answer C. b. is

tangare [24]

The first step in solving quadratic equations by finding square roots is; C:square root both sides to isolate x

<h3>How to solve quadratic equations?</h3>

To answer this question, we will take an example of a quadratic equation that we need to find the square root as;

x² = 36

Now, to get the roots which are the values of x, we will first have to take the square root of both sides to Isolate x. Thus;

√x² = √36

x = ±6

Read more about quadratic equations at; brainly.com/question/1214333

#SPJ1

6 0

1 year ago

Angela shared a cab with her friends. When they arrived at their destination, they evenly divided the

FrozenT [24]

Answer:

$\frac{48+5j}{j}$

Step-by-step explanation:

<u>Complete Question:</u>

Angela shared a cab with her friends . When they arrived at their destination, they evenly divided the $48 fare among "j" of them . Angela gave the cab $5 tip. How much did Angela pay in total?

<u>Solution:</u>

The fare was $48 and including Angela, there were j people, so each paid:

48/j

Also, Angela paid $5 additionally.

So, Angela paid:

48/j --- when dividing evenly

5 --- as a tip

Total Angela Paid: $=\frac{48}{j}+5\\=\frac{48+5j}{j}$

Total Payment: $\frac{48+5j}{j}$

8 0

2 years ago

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Which inequality is the solution to 20 + 2 < x/3 - 8?

pychu [463]

Answer:

x > 90, C

Step-by-step explanation:

20 + 2 < x / 3 - 8

x / 3 - 8 > 22

x / 3 > 30

x > 90

4 0

2 years ago

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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

2 years ago

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