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SIZIF [17.4K]
3 years ago
12

H(x)=-x^2+10x-16 Find the 2 x-intercepts

Mathematics
1 answer:
Katena32 [7]3 years ago
7 0

Answer:

x=2   x=8

Step-by-step explanation:

h(x)=-x^2+10x-16

Set this equal to 0

0=-x^2+10x-16

Divide by -1

0/-1=x^2-10x+16

0=x^2-10x+16

Factor

What two numbers multiply to 16 and add to -10

-2*-8 = 16

-2+-8 = -10

0=(x-2) (x-8)

Using the zero product property

x-2=0   x-8=0

x=2   x=8

These are the x intercepts

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770,070 in expanded form
Shalnov [3]
700,000 + 70,000 + 70


is your answer

hope this helps :D
4 0
3 years ago
Read 2 more answers
Write the equation of the line that passes through the points (-4,-4) and (6,3).
Phantasy [73]
I hope this helps I took both of the points and used the slope formula to get my slope which is 7/10 then I took any point from those two then I plugged it in to get the point slope form

Answer: y-3=7/10(x-6)

6 0
3 years ago
((-2^2)(-1^-3))•((-2^-3)(-1^5))^-2
Vitek1552 [10]

Answer:

  256

Step-by-step explanation:

A calculator works well for this.

_____

None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.

1 to any power is still 1, so the factors (1^n) can be ignored.

After you ignore all of the things that can be ignored, your problem simplifies to ...

  (2^2)(2^-3)^-2

The rules of exponents applicable to this are ...

  (a^b)^c = a^(b·c)

  (a^b)(a^c) = a^(b+c)

Then your product simplifies to ...

  (2^2)(2^((-3)(-2)) = (2^2)(2^6)

  = 2^(2+6)

  = 2^8 = 256

3 0
3 years ago
Question in image again
Reil [10]
16in I believe irdk
Hope it helped..
7 0
3 years ago
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1. When a cannonball is fired, the equation of its pathway can be modeled by h = -1612 + 128t.
sashaice [31]

Answer:

The maximum height of the ball is 256m

Step-by-step explanation:

Given the equation of a pathway modelled as pathway can be modeled by h = -16t² + 128t

At maximum height, the velocity of the ball is zero.

velocity = dh/dt

velocity = -32t + 128

Since v = 0 at maximum height

0 = -32t+128

32t = 128

t = 128/32

t = 4seconds

The maximum height can be gotten by substituting t = 4 into the modelled equation.

h = -16t² + 128t

h = -16(4)²+128(4)

h = -16(16)+512

h = -256+512

h = 256m

4 0
3 years ago
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