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3 years ago

5

Jana is thinking of a number. If she divides her number by 12 and then multiplies the quotient by 8, the result is 520. What num

ber is Jana thinking of?

2 answers:

garri49 [273]3 years ago

8 0

You would do the reverse of what was done so you would do 520/8 and then multiply that by 12 so you get 780

dusya [7]3 years ago

6 0

65 is the answer. because you just divide 520 by 8 and you get 65

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Factors of 42 <br> I need to know what factors make 42

vodomira [7]

<em>1 x 42</em>

<em>2 x 21</em>

<em>3 x 14</em>

<em>6 x 7</em>

Step-by-step explanation:

We need to find factors of 42.

Factors: <em>Two numbers when multiplied if give you the required number then those 2 numbers are factors o that number.</em>

So, Factors of 42 are:

<em>1 x 42</em>

<em>2 x 21</em>

<em>3 x 14</em>

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Only these numbers when multiplied will give 42.

So, All these numbers are factors of 42

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Learn more about factors of numbers at:

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3 years ago

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3 years ago

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3 years ago

If 2 (3t - 10) + t = 40 + 4t, what is the value of 3t?

Elenna [48]

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Step-by-step explanation:

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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