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2 years ago

Give the reason for the steps used in solving the equation 3(x - 4) + 5 = 2x - 2(3 - x)

2 answers:

7 0

$3(x - 4) + 5 = 2x - 2(3 - x) \\ 3x - 12 + 5 = 2x - 6 + 2x \\ 3x - 7 = 4x - 6 \\ 3x - 4x - 7 + 6 = 0 \\ - x - 1 = 0 \\ x = - 1$

Please give brainliest

Triss [41]2 years ago

3 0

Answer:

-1

Step-by-step explanation:

Solve

3x - 7 = 4x - 6

3x - 7 + 7 = 4x - 6 + 7

3x = 4x + 1

3x - 4x = 4x + 1 - 4x

-x • 1

-1

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IfA=−3n+2 andB=5n−7, what is the value of A−B, in simplest form?

alukav5142 [94]

Answer:

Either -4n or 0n but most likely -4n

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3 years ago

Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

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

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

$r=\sqrt{(-2)^{2}+(-3)^{2}}$

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

7 0

3 years ago

W=1000M-200 rearrange your make M the subject

Anton [14]

Answer:

M = $\frac{w+200}{1000}$

Step-by-step explanation:

Given

w = 1000M - 200 ( isolate 1000M by adding 200 to both sides )

w + 200 = 1000M ( divide both sides by 1000 )

$\frac{w+200}{1000}$ = M

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3 years ago

What is the value of k if - 2/3(k - 6) = 3/2(k+6)

Colt1911 [192]

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Answer: -30/13

I hope this helped!

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- Zack Slocum

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3 years ago

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KiRa [710]

Answer:

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Step-by-step explanation:

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3 years ago

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