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larisa [96]
3 years ago
7

Factor the following polynomial: "4x2−16"

Mathematics
2 answers:
ra1l [238]3 years ago
8 0

Answer:

Step-by-step explanation:

does it have to be right?

mario62 [17]3 years ago
4 0
You can take the square root of everything there and it leads to this.

(2x-4)(2x+4)
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Joshua babysits for his sister after school. He is paid
Darina [25.2K]
The answer is 14.50. Hope that helped
4 0
3 years ago
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The sum of two numbers is 65. The larger number is 5 more than the smaller number. What are the numbers?
ipn [44]

Answer:

30 and 35

Step-by-step explanation:

Put this question into equation form:

Let x be the smaller number, if the larger number is 5 more and the total is 65:

x + 5 + x = 65

2x + 5 = 65

Subtract 5 from both sides

2x + 5 - 5 = 65 - 5

2x = 60

Divide both sides by 2:

2x ÷ 2 = 60 ÷ 2

x = 30

As the larger number is 5 bigger:

30 + 5 = 35

So the two numbers are 30 and 35.

Hope this helps!

8 0
2 years ago
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Sooo..... can anybody help.....
Margaret [11]
V= wlh so the answer is 56.25 or 56 1/4
4 0
3 years ago
Read 2 more answers
HELP?
poizon [28]
My "work" is to enter the equations into a graphing calculator and let it show me the answers--just as your "work" is to copy the question to Brainly.

The solutions are
  (x, y) = (-0.5, 7.5)
  (x, y) = (3, 25)

_____
When you equate the expressions for y, you get the quadratic
  2x² -5x -3 = 0
  (2x +1)(x -3) = 0
  x = -1/2 or 3
  y = 5(x +2) = 15/2 or 25

4 0
3 years ago
Lim x-1 x2 - 1/ sin(x-2)
balu736 [363]

Answer:

           \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0

Explanation:

Assuming the correct expression is to find the following limit:

         \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}

Use the property the limit of the quotient is the quotient of the limits:

         \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}

Evaluate the numerator:

          \frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}

Evaluate the denominator:

  • Since         \lim_{x \to1}sin(x-2)\neq 0

                  \frac{0}{\lim_{x \to1}sin(x-2)}=0

4 0
3 years ago
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