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8090 [49]
3 years ago
7

How do i find answes to endjunity geomatry

Mathematics
1 answer:
konstantin123 [22]3 years ago
5 0

Answer:

Hey there !

Under the More button, select View Course Structure.

Find the lesson to view the assessment answers. Click Quiz Answers.

All the assessment questions related to the lesson are found in the pop-up window. To view a question and answer, select a question number.

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What is 3(2x-6)-11=4(x-3)+6
navik [9.2K]

Answer:

x=11.5

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A university is building a new student center that is two- thirds the distance from the arts center to the residential complex.
Natasha2012 [34]

Answer:

C = (\frac{21}{5},\frac{33}{5})

Step-by-step explanation:

Given

Points: (1, 9) and (9, 3)

Ratio = 2/3

Required

Determine the coordinate of the center

Represent the ratio as ratio

Ratio = 2:3

The new coordinate can be calculated using

C = (\frac{mx_2 + nx_1}{n + m},\frac{my_2 + ny_1}{n + m})

Where

(x_1,y_1) = (1, 9)

(x_2, y_2) = (9, 3)

m:n = 2:3

Substitute these values in the equation above

C = (\frac{2 * 9 + 3 * 1}{3 + 2},\frac{2 * 3 + 3 * 9}{2 + 3})

C = (\frac{18 + 3}{5},\frac{6 + 27}{5})

C = (\frac{21}{5},\frac{33}{5})

Hence;

<em>The coordinates of the new center is </em>C = (\frac{21}{5},\frac{33}{5})<em></em>

7 0
3 years ago
Is the expression 7 (6x-8y) equal to 42x-56y
Bond [772]
Yes☺
you are right 7 times 6 is 42 and 7 times. 8 is 56
3 0
3 years ago
Read 2 more answers
Help please!posted picture of question
Evgen [1.6K]
The region is in the first quadrant, and the axis are continuous lines, then x>=0 and y>=0
The region from x=0 to x=1 is below a dashed line that goes through the points:
P1=(0,2)=(x1,y1)→x1=0, y1=2
P2=(1,3)=(x2,y2)→x2=1, y2=3
We can find the equation of this line using the point-slope equation:
y-y1=m(x-x1)
m=(y2-y1)/(x2-x1)
m=(3-2)/(1-0)
m=1/1
m=1
y-2=1(x-0)
y-2=1(x)
y-2=x
y-2+2=x+2
y=x+2
The region is below this line, and the line is dashed, then the region from x=0 to x=1 is:
y<x+2 (Options A or B)

The region from x=2 to x=4 is below the line that goes through the points:
P2=(1,3)=(x2,y2)→x2=1, y2=3
P3=(4,0)=(x3,y3)→x3=4, y3=0
We can find the equation of this line using the point-slope equation:
y-y3=m(x-x3)
m=(y3-y2)/(x3-x2)
m=(0-3)/(4-1)
m=(-3)/3
m=-1
y-0=-1(x-4)
y=-x+4
The region is below this line, and the line is continuos, then the region from x=1 to x=4 is:
y<=-x+2 (Option B)

Answer: The system of inequalities would produce the region indicated on the graph is Option B

8 0
3 years ago
A parabola has vertex (2, 3) and contains the point (0, 0). find an equation that represents this parabola.
AfilCa [17]
First write it in vertex form :-

y= a(x - 2)^2 + 3    where a is some constant.

We can find the value of a  by substituting the point (0.0) into the equation:-

0  = a((-2)^2 + 3

4a = -3

a = -3/4

so our equation becomes  y =  (-3/4)(x - 2)^2 + 3




7 0
3 years ago
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