Factor completely. 7 + 14x + 7x^2 = khan academy

Mathematics

2 answers:

Helen [10]2 years ago

7 0

Answer:

It's 7(1+x)^2

Ipatiy [6.2K]2 years ago

4 0

Answer:

( X + 1 )( X + 1 )

Step-by-step explanation:

just factor it with calculator

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How to find the radius of square that tells you length, width, and height.

Anna [14]

Well diameter will be the diagnol of that square..

let side of square be x, then diagonal will be root2 * x = diameter

so, raidus will be = dia/ 2
= (root2 * x ) /2

5 0

2 years ago

Which mixed number is equivalent to the improper function below? 46/10

Valentin [98]

Answer:

B. 4 and 6/10

Step-by-step explanation:

in order to find this answer i've always had a trick where i first off find an even number to divide these with and in this case that would be 4 and then the leftovers would be 6/10 and in this case your answer would be right there 4 and 6/10 but if needed to be simplified it would be 4 and 3/5

hope this helps, if it did please make sure to like and rate and give me the brainliest if possible!

have a good day!

6 0

2 years ago

Read 2 more answers

What is the answer to this

LekaFEV [45]

$4 \dfrac{3}{5} \times \dfrac{5}{12} =$

$= \dfrac{23}{5} \times \dfrac{5}{12}$

You changed the mixed numeral into a fraction correctly.
Now you need to multiply the fractions.
Multiply the numerators together, and multiply the denominators together.

$= \dfrac{23 \times 5}{5 \times 12}$

Now we could multiply out the numerator and denominator, and then we'd need to reduce the fraction, but before we multiply, we see there is a factor of 5 both in the numerator and in the denominator, so we can reduce before we multiply.

$= \dfrac{23 \times 5}{12 \times 5}$

$= \dfrac{23}{12}$

$= 1 \dfrac{11}{12}$

6 0

2 years ago

If 2/3 of amount x is 60, then what is 4/5 of x

frozen [14]

I think it's 72 then?

8 0

3 years ago

I will give brainliest answer

adell [148]

Answer: A=351.68 cm²

Step-by-step explanation:

To find the area of the shaded region, we would subtract the area of the circle by the area of the inner circle.