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2 years ago

/Algebra/Please Help!?

Mathematics

1 answer:

umka21 [38]2 years ago

5 0

Answer:

answer is b babe

Step-by-step explanation:

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PLEASE HELP WILL MARK BRAINLIEST!!!!!

trapecia [35]

The numbers inside the parenthesis is the value of x, uses the equation associated with the value of x and solve.

H(1) Since 1 fits 1≤x≤3 use x^3

x^3 1^3 = 1

h(1) = 1

h(4) 4 fits x >3, so use 5

h(4) = 5

5 0

2 years ago

Reposting becuase it want let me edit my last one and somebody spam answered it.

harkovskaia [24]

Answer:

5Step-by-step explanation:

If you use the distance formula you can solve it pretty easily. The distance formula is:

d = $\sqrt{(x2 - x1)^2 + (y2 - y1)^2} \\$

The two points you want are A and D. A is (-1, -4) and D is (3, -1). Plug that into the formula...

d = $\sqrt{(3 - (-1))^2 + (-1 - (-4))^2}$

d = $\sqrt{4^2 + 3^2}$

d = $\sqrt{16 + 9}$

d = $\sqrt{25}$

I can solve it in a different way if you want me to.

8 0

3 years ago

Please help i will give brainliest

inysia [295]

Answer:

Step-by-step explanation:

Take the area of the yard and subtract from it the area of the pool. In quadratic form, the area of the pool is

$2x^2-x-28$

Subtracting the area of the pool from the area of the yard:

$3x^2-14x-5-(2x^2-x-28)$

Since the negative in front of the parenthesis will change the signs inside:

$3x^2-14x-5-2x^2+x+28$

Combine like terms to get the area left after the pool goes in:

$x^2-13x+23$

8 0

3 years ago

Factor p^2-8p-20 plz help

Kamila [148]

What 2 numbers multiply to get -20 and add to get -8
-10 and 2

(p-10)(p+2)

3 0

3 years ago

/Algebra/Please Help!?​

/Algebra/Please Help!?