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3 years ago

What is the solution to the equation?

2 answers:

8 0

C. x=3 because you need to distribute the 3 to the x and the -5. This will get you 3x-15= 8x-30. Then, you have to subtract 3x from both sides. -15=5x-30. Lastly, you add 30 to both sides. 15=5x. Divide both sides by 5 and you have x=3.

never [62]3 years ago

7 0

3(x - 5) = 8x - 30

Distribute 3 into the parenthesis:
3x - 15 = 8x - 30

Add 15 to both sides:

3x = 8x - 15

Subtract 8x to both sides:

-5x = -15

Divide -5 to both sides:

x = 3

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3 years ago

Someone please be awesome and help me please :(

solong [7]

Answer:

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Step-by-step explanation:

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

Now we are read to write that one part (the first three terms together) as a square:

$(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

$(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0$

They put it in ( )

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

I'm going to go ahead and combine those fractions now:

$(x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a^2})=0$

I'm going to factor out a -1 in the second term ( the one in the second ( ) ):

$(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0$

Now I'm going to add (b^2-4ac)/(4a^2) on both sides:

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

I'm going to square root both sides to rid of the square on the x+b/(2a) part:

$x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$

$x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}$

Now subtract b/(2a) on both sides:

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Combine the fractions (they have the same denominator):

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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There’s no picture, please post one

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3 years ago

What are the domain and range of this function?

RUDIKE [14]

<h3>Answer: Choice D</h3>

Domain: all real numbers
Range: $y \ge 1$

===========================================

Explanation:

The domain is the set of allowed x inputs. We can plug in any x value we want as the graph stretches on forever to the left and to the right. There aren't any division by zero errors or any issues like that to worry about, so that's why we don't kick out any x values from the domain.

The domain being all real numbers translates to the interval notation $(-\infty, \infty)$ which is basically saying $-\infty < x < \infty$

-----------------------------------

The range is the set of y outputs possible. The graph shows that y = 1 is the smallest y output, so y = 1 or y can be greater than this.

In short, $y \ge 1$ is the range which converts to the interval notation $[1, \infty)$ which is the same as saying $1 \le y < \infty$

-----------------------------------

Extra info:

The equation of this absolute value function is y = |x|+1

8 0

2 years ago