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3 years ago

The sum of two numbers is eleven. One number is thirty-eight less than six times the other. Find the numbers.

Mathematics

1 answer:

snow_lady [41]3 years ago

3 0

Answer:

One of the number is 4 while the other is 7

Step-by-step explanation:

The sum of two numbers is eleven

Let the numbers be x and y

X+ y = 11

One number is thirty-eight less than six times the other

X= 6y-38

X+ y = 11

6y-38 +y= 11

6y+y= 11+38

7y=49

Y= 49/7

Y= 7

X+ y = 11

X+7= 11

X= 11-7

X= 4

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blagie [28]

Greetings and Happy Holidays!

Solve for n.
$4(3n-2)^{\frac{-3}{2}}+1= \frac{347}{343}$

Add -1 to both sides.
$(4(3n-2)^{\frac{-3}{2}}+1)+(-1)=(\frac{347}{343})+(-1)$

$4(3n-2)^{\frac{-3}{2}}=\frac{347}{343}-1$

$4(3n-2)^{\frac{-3}{2}}=\frac{347}{343}-\frac{343}{343}$

$4(3n-2)^{\frac{-3}{2}}=\frac{4}{343}$

Divide both sides by 4
$\frac{4(3n-2)^{\frac{-3}{2}}}{4}=\frac{\frac{4}{343}}{4}$

$\frac{4(3n-2)^{\frac{-3}{2}}}{4}=\frac{4}{343}(\frac{1}{4})$

$\frac{4(3n-2)^{\frac{-3}{2}}}{4}=\frac{4}{1372}$

$\frac{4(3n-2)^{\frac{-3}{2}}}{4}=\frac{1}{343}$

$(3n-2)^{\frac{-3}{2}}=\frac{1}{343}$

Solve the Exponent.
$((3n-2)^{\frac{-3}{2}})^{\frac{2}{-3}} =\frac{1}{343}^{\frac{2}{-3}}$

$3n-2=49$

Add 2 to both sides.
$(3n-2)+2=(49)+2$

$3n=51$

Divide both sides by 3.
$\frac{3n}{3}= \frac{51}{3}$

$n=17$

The Answer Is:
$\left[\begin{array}{ccc}n=17\end{array}\right]$

Hope this helped!
-Benjamin

5 0

3 years ago