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Mrac [35]
2 years ago
9

Find the range of the function. ƒ(x) = x2 + 3

Mathematics
1 answer:
beks73 [17]2 years ago
7 0

Answer:

c

Step-by-step explanation:

You might be interested in
6-2(x+6)=3x+4 <br><br><br> ...............................
juin [17]

Answer:

x=−2

Step-by-step explanation:

1 Expand.

6-2x-12=3x+4

6−2x−12=3x+4

2 Simplify  6-2x-126−2x−12  to  -2x-6−2x−6.

-2x-6=3x+4

−2x−6=3x+4

3 Add 2x2x to both sides.

-6=3x+4+2x

−6=3x+4+2x

4 Simplify  3x+4+2x3x+4+2x  to  5x+45x+4.

-6=5x+4

−6=5x+4

5 Subtract 44 from both sides.

-6-4=5x

−6−4=5x

6 Simplify  -6-4−6−4  to  -10−10.

-10=5x

−10=5x

7 Divide both sides by 55.

-\frac{10}{5}=x

−

5

10

​

=x

8 Simplify  \frac{10}{5}

5

10

​

  to  22.

-2=x

−2=x

9 Switch sides.

x=-2

x=−2

4 0
2 years ago
HELP RN PLEASE.
Viktor [21]

Answer:

By showing that readers know more about the rebellion that most of the animals

Step-by-step explanation:

The answer is correct on edg2020

6 0
2 years ago
Read 2 more answers
Name the correct angles 5 and 7
Alex787 [66]

Answer:

Acute angle.

Right angle.

Obtuse angle.

Straight angle.

Reflex angle

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Integrate [0, 1/2] xcos(pi*x).
elena-14-01-66 [18.8K]
To answer your question, this could be the possible answer and i hope you understand and interpret it correctly:
<span>[Integrate [0, 1/2] xcos(pi*x
let u=x so that du=dx
 and v=intgral cos (xpi)dx
        v=(1/pi)sin(pi*x)
 integration by parts
uv-itgral[0,1/2]vdu just plug ins
 (1/pi)sinpi*x]-(1/pi)itgrlsin(pi*x)dx from 0 to 1/2
(1/pi)x sinpi*x - (1/pi)[-(1/pi) cos pi*x] from 0 to 1/2
 =(1/2pi)+(1/pi^2)[cos pi*x/2-cos 0]
=1/2pi - 1/2pi^2
=(pi-2)/2pi^2 ans</span>
3 0
3 years ago
Can you please explain and answer this question to me, (don't mind the blue dots, they are the answers of what I think, please l
Fofino [41]
I started with c, but you could’ve started with any box.

So I took the numbers in the c box (the black box in my picture) and divided to see which were multiples of two. They all were.

Then I went to a. All of a’s numbers were divisible by 7.

Then b. Two of them are divisible by 2, so that’s not the answer. None of them were divisible by 7, so there’s your answer!

Then d. Two are divisible by 7, so your rules is divisible by 2.

I hope this helps!

6 0
3 years ago
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