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gladu [14]
3 years ago
14

Explain why the equation (x-3)^2-38=11 has two solutions. Then solve the equation to find the solutions. Show your work.

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
4 0

Answer:

X = 10 and X = -4

Step-by-step explanation:

(x-3)^2-38=11

(x-3)^2 = 11 +38

(x-3) ^2 = 49

Taking square root on both sides

√(x-3) ^2 = √ 49

X-3 = +7, -7

X = 7 +3,    X= -7+3

X = 10        X = -4

So, it has got two solutions

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Wirte a decimal for each of the following 9x100+2x10+3x0.1+7x0.001
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Given f(x) = -3x - 2, find the following:<br> a. f(3)<br> b. f(-1)<br> c. f(-2)
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2 years ago
SOMEONE HELP MEEEEEE 75 POINTS TO THE PERSON THAT HELPS
Tresset [83]

Answer:

Part 1) 9x-7y=-25

Part 2) 2x-y=2

Part 3) x+8y=22  

Part 4) x+8y=35

Part 5) 3x-4y=2

Part 6) 10x+6y=39

Part 7) x-5y=-6

Part 8)

case A) The equation of the diagonal AC is x+y=0

case B) The equation of the diagonal BD is x-y=0

Step-by-step explanation:

Part 1)

step 1

Find the midpoint

The formula to calculate the midpoint between two points is equal to

M=(\frac{x1+x2}{2},\frac{y1+y2}{2})

substitute the values

M=(\frac{2-6}{2},\frac{-3+5}{2})

M=(-2,1)

step 2

The equation of the line into point slope form is equal to

y-1=\frac{9}{7}(x+2)\\ \\y=\frac{9}{7}x+\frac{18}{7}+1\\ \\y=\frac{9}{7}x+\frac{25}{7}

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

Ax+By=C

where

A is a positive integer, and B, and C are integers

y=\frac{9}{7}x+\frac{25}{7}

Multiply by 7 both sides

7y=9x+25

9x-7y=-25

Part 2)

step 1

Find the midpoint

The formula to calculate the midpoint between two points is equal to

M=(\frac{x1+x2}{2},\frac{y1+y2}{2})

substitute the values

M=(\frac{1+5}{2},\frac{0-2}{2})

M=(3,-1)

step 2

Find the slope

The slope between two points is equal to

m=\frac{-2-0}{5-1}=-\frac{1}{2}

step 3

we know that

If two lines are perpendicular, then the product of their slopes is equal to -1

Find the slope of the line perpendicular to the segment joining the given points

m1=-\frac{1}{2}

m1*m2=-1

therefore

m2=2

step 4

The equation of the line into point slope form is equal to

y-y1=m(x-x1)

we have

m=2 and point (1,0)

y-0=2(x-1)\\ \\y=2x-2

step 5

Convert to standard form

Remember that the equation of the line into standard form is equal to

Ax+By=C

where

A is a positive integer, and B, and C are integers

y=2x-2

2x-y=2

Part 3)

In this problem AB and BC are the legs of the right triangle (plot the figure)

step 1

Find the midpoint AB

M1=(\frac{-5+1}{2},\frac{5+1}{2})

M1=(-2,3)

step 2

Find the midpoint BC

M2=(\frac{1+3}{2},\frac{1+4}{2})

M2=(2,2.5)

step 3

Find the slope M1M2

The slope between two points is equal to

m=\frac{2.5-3}{2+2}=-\frac{1}{8}

step 4

The equation of the line into point slope form is equal to

y-y1=m(x-x1)

we have

m=-\frac{1}{8} and point (-2,3)

y-3=-\frac{1}{8}(x+2)\\ \\y=-\frac{1}{8}x-\frac{1}{4}+3\\ \\y=-\frac{1}{8}x+\frac{11}{4}

step 5

Convert to standard form

Remember that the equation of the line into standard form is equal to

Ax+By=C

where

A is a positive integer, and B, and C are integers

y=-\frac{1}{8}x+\frac{11}{4}

Multiply by 8 both sides

8y=-x+22

x+8y=22  

Part 4)

In this problem the hypotenuse is AC (plot the figure)

step 1

Find the slope AC

The slope between two points is equal to

m=\frac{4-5}{3+5}=-\frac{1}{8}

step 2

The equation of the line into point slope form is equal to

y-y1=m(x-x1)

we have

m=-\frac{1}{8} and point (3,4)

y-4=-\frac{1}{8}(x-3)

y=-\frac{1}{8}x+\frac{3}{8}+4

y=-\frac{1}{8}x+\frac{35}{8}

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

Ax+By=C

where

A is a positive integer, and B, and C are integers

y=-\frac{1}{8}x+\frac{35}{8}

Multiply by 8 both sides

8y=-x+35

x+8y=35

Part 5)  

The longer diagonal is the segment BD (plot the figure)  

step 1

Find the slope BD

The slope between two points is equal to

m=\frac{4+2}{6+2}=\frac{3}{4}

step 2

The equation of the line into point slope form is equal to

y-y1=m(x-x1)

we have

m=\frac{3}{4} and point (-2,-2)

y+2=\frac{3}{4}(x+2)

y=\frac{3}{4}x+\frac{6}{4}-2

y=\frac{3}{4}x-\frac{2}{4}

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

Ax+By=C

where

A is a positive integer, and B, and C are integers

y=\frac{3}{4}x-\frac{2}{4}

Multiply by 4 both sides

4y=3x-2

3x-4y=2

Note The complete answers in the attached file

Download docx
3 0
3 years ago
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