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STatiana [176]
3 years ago
9

Given the function f(x)=6 (4x-6)+18, determine t e value of x such that f(x)=6

Mathematics
2 answers:
In-s [12.5K]3 years ago
8 0
I hope this helps you!

notsponge [240]3 years ago
5 0

Answer: x=1

Step-by-step explanation:

Prepworks;))

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We want to find the zeros of this polynomial:
slava [35]

Answer:

x=-3        x=-1            x=2         x=-2

Step-by-step explanation:

p(x) = (x^2 + 4x + 3)(x^2 – 4)

Set this equal to zero to find the x intercepts

0 = (x^2 + 4x + 3)(x^2 – 4)

Using the zero product property

(x^2 + 4x + 3) =0      (x^2 – 4)  =0

Factor

(x+3)(x+1) =0            (x-2) (x+2)=0

Using the zero product property

x+3 =0  x+1 =0          x-2 =0  x+2 =0

x=-3        x=-1            x=2         x=-2

7 0
3 years ago
Jackson wants to tile the floor in his bathroom with square tiles that are 1/2 foot long. Will he use more or fewer tiles if he
andreev551 [17]
He will use more tiles because 1/3 is smaller than 1/2 and the smaller the tiles the more you need to fill the place up.
3 0
3 years ago
Which of these is the best definition of an ellipse?
MArishka [77]

Answer:

I think it's B.

Step-by-step explanation:

Sorry if I'm wrong :/

4 0
3 years ago
Read 2 more answers
Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)
Mnenie [13.5K]
One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
(x-h)^2+(y-k)^2=r^2
the center is (h,k) and the radius is r

and the distance formula is
distance between (x_1,y_1) and (x_2,y_2) is
D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}


r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle


so
(x+2)^2+(y-3)^2=18
(x-(-2))^2+(y-3)^2=(\sqrt{18})^2
(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2

the radius is 3\sqrt{2}
center is (-2,3)

find distance between (8,4) and (-2,3)

D=\sqrt{(8-(-2))^2+(4-3)^2}
D=\sqrt{(8+2)^2+(1)^2}
D=\sqrt{10^2+1}
D=\sqrt{100+1}
D=\sqrt{101}




r=3\sqrt{2}≈4.2
D=\sqrt{101}≈10.04

do r<D

(8,4) is outside the circle

6 0
3 years ago
PLEASE HELP i’ll mark brainliest ASAPP
chubhunter [2.5K]

Answer:

C. 4x squared -36

Step-by-step explanation:

8 0
3 years ago
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