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3 years ago

Which table of values includes solutions to the function f(x) = -2x^2?

Mathematics

1 answer:

olga_2 [115]3 years ago

3 0

The table could include values like this:
x y
-2 -8
-1 -2
0 0
1 -2
2 -8

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Find the point of the intersection for the given systems y=x^2+1 and y=x+1

mario62 [17]

set them equal to each other

x²+1=x+1

minus 1 from both sides

x²=x

minus x both sides

x²-x=0

factor

x(x-1)=0

set each to 0

x=0

x-1=0

x=1

subsitute back to find y values

y=x+1, y=0+1, y=1, one point is (0,1)

y=x+1, y=1+1, y=2, another point is (1,2)

the 2 points of intersection are (0,1) and (1,2)

7 0

3 years ago

Please help me please I need help!!!!!!!!!!!!!!!!!!!hshdjdndndbdbnfjfkfkfkdkdkfkfjdjfiid

Serggg [28]

y = (x - 2)(x + 5) First I would multiply (x - 2) and (x + 5) together

y = x² + 5x - 2x - 10

y = x² + 3x - 10

Next I would plug in a number for x to find its y value. I will plug in 0

y = 0² + 3(0) - 10

y = -10

(0, -10) [this is the y-intercept (the y value when x = 0)]

Your answer is Graph A because the y-intercept is (0,-10)

6 0

3 years ago

Liang enlarged a photograph to 1 3/4 times it original size. The original photograph was 5 inches wide. What was the width of th

Alja [10]

8.75 inches
5 divided by 4 is 1.25
1.25 x 3 = 3.75
3.75 + 5 = 8.75

7 0

3 years ago

Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

Put $(x,y)=(1,1)$

$\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9$

This is true.

So, the point $(1,1)$ satisfies the equation $\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

So, point H is $(1,1)$ .

7 0

3 years ago

What number make the sentence true? (7+4)+6( +6

Arte-miy333 [17]

Was this typed right????????????

6 0

3 years ago