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3 years ago

Complete the function for this graph.

Mathematics

1 answer:

ikadub [295]3 years ago

7 0

Answer:

y = –|x - 1| + 3

Step-by-step explanation:

The "vertex" is (1 , 3)

y = –|x - 1| + 3

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The diameter of a ball is 8 in.

sergejj [24]

Answer:

267.95

Step-by-step explanation:

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2 years ago

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Is 2×3×4the prime factorization of 24 Explain you reasoning

Elden [556K]

Answer:

see the prime factors

Step-by-step explanation:

24= 6*4=3*2*2*2

=2*3*4

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3 years ago

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X-intercept=2; containing the point (-5,4), using either the general form or the slope-intercept form of the equation of a line.

Rus_ich [418]

The line contains point (2, 0) and (-5, 4).
y - 0 = (4 - 0)/(-5 - 2) (x - 2)
y = -4/7 (x - 2)
7y = -4(x - 2)
7y = -4x + 8
4x + 7y = 8

8 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

2 years ago

Given the equation 5x − 4 = –2(3x + 2), solve for the variable. Explain each step and justify your process.

blagie [28]

A.
$5x-4=-2(3x+2) \ \ \ \ \ \ \ \ \ |\hbox{expand the bracket} \\
5x-4=-2 \times 3x-2 \times 2 \\
5x-4=-6x-4 \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 6x to both sides} \\
11x-4=-4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 4 to both sides} \\
11x=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{divide both sides by 11} \\
x=0$

B.
$3(2x-4)=5x-1 \\
6x-12=5x-1 \\
\boxed{11x-12=-1} \Leftarrow \hbox{the first mistake} \\
11x=11 \\
\boxed{x=11} \Leftarrow \hbox{the second mistake}$

Megan's solution isn't correct.
The first mistake: she subtracted 5x from the right-hand side of the equation, but added 5x to the left-hand side.
The second mistake: she divided the right-hand side of the equation by 11, but didn't divide the left-hand side.

The correct solution:
$3(2x-4)=5x-1 \ \ \ \ \ \ \ \ \ \ \ |\hbox{expand the bracket} \\
3 \times 2x+3 \times (-4)=5x-1 \\ 6x-12=5x-1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{subtract 5x from both sides} \\
x-12=-1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\hbox{add 12 to both sides} \\
x=11$

3 0

2 years ago