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2 years ago

What is the solution to 4x + 2 = 6(-2x - 5) ?

Mathematics

2 answers:

exis [7]2 years ago

5 0

Answer:

x= -2

Step-by-step explanation:

4x + 2 = 6(-2x - 5)

Distribute

4x+2 = -12x-30

Add 12x to each side

4x+2+12x = -12x-30+12x

16x+2 = -30

Subtract 2 from each side

16x+2-2=-30-2

16x = -32

Divide by 16

16x/16x = -32/16

x = -2

Luba_88 [7]2 years ago

4 0

The answer is x=-2 !!!!!!!!

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The answer to<br> 6+2(17-13)²

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Answer:

Step-by-step explanation:

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3 years ago

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F(x) = 3x + 2 and g(x) = x + 5 g[f(x)] =

Tema [17]

Answer:

f (2x+1)=6x−2

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3 years ago

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Identify the equation in slope-intercept for for the line containing the point (-3,5) and patella to y = -2/3x+5/3

Olegator [25]

Answer:

Y = - 2/3 x + 3

Step-by-step explanation:

y = -2/3x+5/3

Slope = -2/3. .. parallel line has same slope

For (- 3 , 5)

Y intercept = y - my = 5 - ( (-2/3) x (-3) ) = 3

Y = slope x X + y intercept

Y = - 2/3 x + 3

8 0

3 years ago

Andrea, Justin, and Virginia went to an office supply store. Andrea bought 6 pencils, 7 markers, and 8 erasers. Her total was $1

olga2289 [7]

Answer:

$x=0.25\,,\,y=1.75\,,\,z=0.25$

Step-by-step explanation:

Let $x,y,z$ denotes number of pencils, markers, erasers respectively.

For Andrea:

$6x+7y+8z=15.75\,\,\,(i)$

For Justin:

$6x+8y+5z=16.75\,\,\,(ii)$

For Virginia:

$5x+5y+7z=11.75\,\,\,(iii)$

On subtracting equations (i) and (ii), we get

$(6x+7y+8z)-(6x+8y+5z)=15.75-16.75\\-y+3z=-1\\y-3z=1\\y=1+3z$

Put $y=1+3z$ in equation (i)

$6x+7(1+3z)+8z=15.75\\6x+29z=8.75\,\,\,(iv)$

Put $y=1+3z$ in equation (iii)

$5x+5(1+3z)+7z=11.75\\5x+22z=6.75\,\,\,(v)$

Multiply equation (iv) by 5 and equation (v) by 6 and then subtract both the equations.

$5(6x+29z)-6(5x+22z)=5(8.75)-6(6.75)\\13z=3.25\\z=\frac{3.25}{13}=0.25$

Put $z=0.25$ in equation $y=1+3z$

$y=1+3(0.25)=1.75$

Put $y=1.75\,,\,z=0.25$ in equation (i)

$6x+7(1.75)+8(0.25)=15.75\\6x+12.25+2=15.75\\6x=1.5\\x=0.25$

8 0

3 years ago

Choose the domain for which each function is defined

klemol [59]

Answer:

Part 1) $f(x)=\frac{x+4}{x}$ -----> $x\neq 0$

Part 2) $f(x)=\frac{x}{x+4}$ ----> $x\neq -4$

Part 3) $f(x)=x(x+4)$ ----> All real numbers

Part 4) $f(x)=\frac{4}{x^2+8x+16}$ ----> $x\neq -4$

Step-by-step explanation:

we know that

The domain of a function is the set of all possible values of x

Part 1) we have

$f(x)=\frac{x+4}{x}$

we know that

In a quotient the denominator cannot be equal to zero

For the value of x=0 the function is not defined

therefore

The domain is

$x\neq 0$

Part 2) we have

$f(x)=\frac{x}{x+4}$

we know that

In a quotient the denominator cannot be equal to zero

For the value of x=-4 the function is not defined

therefore

The domain is

$x\neq -4$

Part 3) we have

$f(x)=x(x+4)$

Applying the distributive property

$f*(x)=x^2+4x$

This is a vertical parabola open upward

The function is defined by all the values of x

therefore

The domain is all real numbers

Part 4) we have

$f(x)=\frac{4}{x^2+8x+16}$

we know that

In a quotient the denominator cannot be equal to zero

Equate the denominator to zero

$x^2+8x+16=0$

Remember that

$x^2+8x+16=(x+4)^2$

( $x+4)^2=0$

The solution is x=-4

For the value of x=-4 the function is not defined

therefore

The domain is

$x\neq -4$

6 0

3 years ago