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oksano4ka [1.4K]
3 years ago
7

Find the number that makes the ratio equivalent to 12:1. ?:6

Mathematics
1 answer:
Naily [24]3 years ago
5 0
Right side was multiplied by 6 so you multiply the left side by 6 so 72:6
You might be interested in
How do you solve 1/3 of 3/4 <br> of a fraction
Trava [24]

1 1/12

The fractions have unlike denominators. First, find the Least Common Denominator and rewrite the fractions with the common denominator.

LCD(1/3, 3/4) = 12

Multiply both the numerator and denominator of each fraction by the number that makes its denominator equal the LCD. This is basically multiplying each fraction by 1.

(13×44)+(34×33)=?

Complete the multiplication and the equation becomes

4/12+9/12=?

The two fractions now have like denominators so you can add the numerators.

Then:

4+9/12=13/12

This fraction cannot be reduced.

The fraction

13/12

is the same as

13÷12

Convert to a mixed number using

long division for 13 ÷ 12 = 1R1, so

1312=1 1/12

Therefore:

13+34=1 11/2

5 0
3 years ago
Solve for x in the equation 2x^2+3x-7=x^2+5x+39
Shalnov [3]
Hey there, hope I can help!

\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}
2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
x^2-2x-46=0

Lets use the quadratic formula now
\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}
x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}

\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

Multiply the numbers 2 * 1 = 2
\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}

\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \  \left(-2\right)^2=2^2, 2^2 = 4

\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \  \sqrt{4+184} \ \textgreater \  \sqrt{188} \ \textgreater \  2 + \sqrt{188}
\frac{2+\sqrt{188}}{2} \ \textgreater \  Prime\;factorize\;188 \ \textgreater \  2^2\cdot \:47 \ \textgreater \  \sqrt{2^2\cdot \:47}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \  \sqrt{47}\sqrt{2^2}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \  \sqrt{2^2}=2 \ \textgreater \  2\sqrt{47} \ \textgreater \  \frac{2+2\sqrt{47}}{2}

Factor\;2+2\sqrt{47} \ \textgreater \  Rewrite\;as\;1\cdot \:2+2\sqrt{47}
\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \  2\left(1+\sqrt{47}\right) \ \textgreater \  \frac{2\left(1+\sqrt{47}\right)}{2}

\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \  1+\sqrt{47}

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  2-\sqrt{188} \ \textgreater \  \frac{2-\sqrt{188}}{2}

\sqrt{188} = 2\sqrt{47} \ \textgreater \  \frac{2-2\sqrt{47}}{2}

2-2\sqrt{47} \ \textgreater \  2\left(1-\sqrt{47}\right) \ \textgreater \  \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \  1-\sqrt{47}

Therefore our final solutions are
x=1+\sqrt{47},\:x=1-\sqrt{47}

Hope this helps!
8 0
2 years ago
Read 2 more answers
How to solve 6x + y= -5, -6x + 12y=96 using elimination and does it have no solution or many solutions?
Alika [10]

Answer:

x = -2

Step-by-step explanation:

Arrange 6x + y= -5, -6x + 12y=96 vertically:

6x + y= -5

-6x + 12y=96       Now combine like terms:

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

       13y = 91

Solving for y by dividing both sides by 13:  y = 7

Find x by substituting 7 for y in the second equation:  -6x + 12(7) = 96

Then -6x = -84 + 96, or:

-6x = 12, and so x = -2

8 0
2 years ago
Find slope of -2,3 and 4,-5
bulgar [2K]
-2,3 is (x1,y1) and (4,-5) is (x2,y2) so you do y2-y1/x2-x1

y2-y1 is -5-3 which is -8 and x2-x1 is 4-(-2) which is 6 so you get

-8/6= -4/3
4 0
2 years ago
1/4 (20x−28) &lt;2(x−3)+2
xeze [42]

Answer:

Step-by-step explanation:

1/4 x 20x - 1/4 x 28<2x - 6 + 2

5x - 28/4<2x - 4

5x - 2x<-4 + 28/4

3x<-4/1 + 28/4

3x<-16 + 28/4

3x<12/4

3x<3

3x/3<3/3

x<1

6 0
2 years ago
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