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Y_Kistochka [10]

3 years ago

8

What is the sum?

xFormula1" title=" \frac{2}{x^2} + \frac{4}{x^2} " alt=" \frac{2}{x^2} + \frac{4}{x^2} " align="absmiddle" class="latex-formula">

1 answer:

Varvara68 [4.7K]3 years ago

7 0

Since you have a common denominator, the $x^{2}$ is gonna be the on the bottom. You'll just have to add 2 + 4, and then put it over your $x^{2}$ .

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Thank you so much for helping meeeeee

Ivenika [448]

Hello, and thanks for posting your question!

A function would be a set of ordered pairs that have no repeating "x" values, look at the example below. (V)

{(1, 0), (2, 0), (3, 0)} = Function
{(1, 0), (1, 0), (2, 0)} = Not a function.

According to this information, your answer is C. Both Set A and Set B.

Hope this helps! ☺♥

7 0

3 years ago

A test consists of 20 problems and students are told to answer any 10 of these questions. In how many different ways can they ch

Furkat [3]

We must find UNIQUE combinations because choosing a,b,c,d... is the same as d,c,b,a...etc. For this type of problem you use the "n choose k" formula...

n!/(k!(n-k)!), n=total number of choices available, k=number of choices made..

In this case:

20!/(10!(20-10)!)

20!/(10!*10!)

184756

3 0

3 years ago

Read 2 more answers

The equation below describes a parabola. If a is positive, which way does the

Rzqust [24]

B. Up

Because a positive ‘a’ value will make the parabola open upwards, and because the power is 2 (even) both sides will open up in a ‘U’ shape

8 0

1 year ago

Read 2 more answers

HELP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE!

kiruha [24]

Answer:

b

i did this before xD

6 0

3 years ago

1011+111 in binary form

Ann [662]

Answer:

10010

Step-by-step explanation:

$1011=1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$111=1(2)^2+1(2)^1+1(2)^0$

So $1011+111$ gives us:

$1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$+$

$1(2)^2+1(2)^1+1(2)^0$

-----------------------------------------------------

Combine like terms:

$1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0$

$1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0$

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace $2^0$ with 1 and $2$ with $(2)^1$ :

$1(2)^3+1(2)^2+(2)^2+(2)^1(1)$

I want a $2^0$ number:

$1(2)^3+1(2)^2+1(2)^2+1(2)^1+0(2)^0$

Combine like terms:

$1(2)^3+2(2)^2+1(2)^1+0(2)^0$

$2(2)^2=2^3$ :

$1(2)^3+2^3+1(2)^1+0(2)^0$

Combine like terms:

$2(2)^3+1(2)^1+0(2)^0$

We can rewrite the first term by law of exponents:

$2^4+1(2)^1+0(2)^0$

$1(2)^4+1(2)^1+0(2)^0$

So the binary form is:

$10010$

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

1 0 1 1

+ 1 1 1

------------------------------

(1) (1) (1)

1 0 1 1

+ 1 1 1

------------------------------

1 0 0 1 0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

8 0

3 years ago