All categories

3 years ago

FOR MY TEST!!! ASAP HELP!!!! WILL MARK THE BRAINLIST WHOEVER GETS IT RIGHT!!!!

Mathematics

2 answers:

scoundrel [369]3 years ago

7 0

The answer is attached.

svp [43]3 years ago

6 0

Answer:

the 2 images are the solutions to the questions

You might be interested in

Bryson hopes to win a three-day vacation in a drawing that is being held at his office. He purchased 40 raffle tickets. There we

Savatey [412]

The correct answer would be A, 2/25.

9 0

3 years ago

Which one is correct please hurry

nikklg [1K]

Answer:

Step-by-step explanation:

it is g because you have to acally read that question

4 0

3 years ago

How to solve this. please help me ASAP. vector chapter

Dafna1 [17]

Answer:

1) m = -1

2) k = -2; h = -8

Step-by-step explanation:

................

8 0

3 years ago

4. See if you can find "like-terms" in the expression and

WITCHER [35]

Answer:

After finding like terms and solving, we get $\mathbf{3x^2+3x+16}$

Step-by-step explanation:

We need to find like terms and simplify the expression $4- 2x + x^2 + 5x + 12 + 2x^2$

Like terms: The terms who have same variable and exponent are called like terms.

So, Combining like terms and solving:

$4- 2x + x^2 + 5x + 12 + 2x^2\\=4+12-2x+5x+x^2+2x^2\\Now \ solving:\\=16+3x+3x^2\\Rearranging \ the \ terms:\\=3x^2+3x+16$

So, after finding like terms and solving, we get $\mathbf{3x^2+3x+16}$

<u><em>Note: In the question given we have - and + sign with the term 5x</em></u> i.e 4- 2x + x^2 -+ 5x + 12 + 2x^2

I have solved using +5x, but if the term is -5x then the solution will be:

$4- 2x + x^2 - 5x + 12 + 2x^2\\=4+12-2x-5x+x^2+2x^2\\Now \ solving:\\=16-7x+3x^2\\Rearranging \ the \ terms:\\=3x^2-7x+16$

So, the answer will be: $\mathbf{3x^2-7x+16}$

7 0

3 years ago

For the concentric circles, the outer circle radius is r=6 in while that of the inner circle is r=3 in. Find the exact area of t

worty [1.4K]

Answer:

The exact area of the annulus (ring) made by the two circle is $27\pi \ in^{2}$

Step-by-step explanation:

Given:

Let the radius of outer circle i.e CA be $r_{o}= 6\ in$

Let the radius of inner circle i.e CB be $r_{i}= 3\ in$

The diagram is given below as attachment.

$\textrm{area of circle}= \pi r^{2} \\\textrm{area of the shaded region} =\textrm{area of outer circle}-\textrm{area of inner circle}\\\textrm{area of the annulus ring}=\pi r_{o}^{}2 - \pi r_{i}^{}2$

Substituting the values we get

$\textrm{area of the annulus ring}=\pi\times 6^{2} - \pi\times 3^{2}\\=\pi (36-9)\\=27\pi\ in^{2}$

4 0

3 years ago