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3 years ago

2-(3x-4)=2x-9 Please give step by step answer!! Thanks!

Mathematics

2 answers:

kirza4 [7]3 years ago

7 0

Simplify both sides of the equation

2-(3x-4)=2x-9

2+-1(3x-4)=2x-9

Distribute the negative sign

2+-1(3x)+(-1)(-4)=2x-9

2+-3x+4=2x-9

2+-3x+4=2x+-9

Combine like terms

(-3x)+(2+4)=2x-9

-3x+6=2x-9

Subtract 2x from both sides

-3x+6-2x=2x-9-2x

-5x+6=-9

Subtract 6 from both sides

-5x+6-6=-9-6

-5x=-15

Divide both sides by -5

-5x/5=-15/-5

x=3

I hope that's help !

egoroff_w [7]3 years ago

4 0

First put the -1 into the 3x and -4, so it looks like 2-3x+4=2x-9, then combine 2 and 4, (-3x+6=2x-9), then subtract 2x from both sides (-5x+6=-9) then subtract 6 to both sides so -5x=-15, then divide by 5 so x=3 hope this helps!! sorry i messed up the first time

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A number, y, is equal to twice the sum of a smaller number and 3. The larger number is also equal to 5 more than 3 times the sma

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Answer:

The equations represent the situation is $y=2x+6$ and $y=5+3x$ .

Step-by-step explanation:

Consider the provided information.

First convert the statement into mathematical representation.

Consider the smaller number is x and the larger number is y.

y, is equal to twice the sum of a smaller number and 3.

This information can be written as: $y=2(x+3)$

Which can be further solve as:

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The larger number is also equal to 5 more than 3 times the smaller number.

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Thus, the equations represent the situation is $y=2x+6$ and $y=5+3x$ .

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

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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

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