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3 years ago

The difference of twice a number and five is three. Find the number.

Mathematics

1 answer:

ss7ja [257]3 years ago

5 0

let x = a number

difference means that its subtraction

twice a number is 2x

so the equation should be

2x-5=3

add 5 to both sides

2x=8

divide both sides by 2

x=4

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Can someone help me out on this please having trouble on both and show work please !!

inysia [295]

The difference of the expression is as follows:

(6y⁴ + 3y² - 7) - (12y⁴ - y² + 5) = - 6y⁴ + 4y² - 12

3(x - 5) - (2x + 4) = x - 19

<h3>How to find the difference of the expression?</h3>

The difference of the expression can be found when we combine the like terms.

Therefore,

(6y⁴ + 3y² - 7) - (12y⁴ - y² + 5)

6y⁴ + 3y² - 7 - 12y⁴ + y² - 5

6y⁴ - 12y⁴ + 3y² + y² - 7 - 5

- 6y⁴ + 4y² - 12

3(x - 5) - (2x + 4)

3x - 15 - 2x - 4

3x - 2x - 15 - 4

x - 19

learn more on expression here: brainly.com/question/14294918

#SPJ1

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2 years ago

So I keep getting 188 or 88 but I got the reminder right I’m not sure what I did wrong

9966 [12]

Answer:

Step-by-step explanation:

7619 / 7

= 1088 remainder 3.

3 0

3 years ago

What is a unit rate?

Ksivusya [100]

Answer:

<em>A unit rate is something over a unit of something else. </em>

<u>Such as:</u> (50 miles) / hour

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3 years ago

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jekas [21]

Answer:

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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):

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

6 0

3 years ago