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3 years ago

-13 times the difference between a number and 60 is equal to the number plus -59

1 answer:

8 0

Let n represent the number.
-13×(n -60) = n +(-59)
-13n +780 = n -59
839 = 14n
839/14 = n = 59 13/14

The number is 59 13/14.

_____
Check
The difference between the number and 60 is
59 13/14 -60 = -1/14
-13 times that value is
-13*(-1/14) = 13/14
Compare that to the number plus -59
59 13/14 -59 = 13/14 . . . . . the same value, as required.

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emmainna [20.7K]

Answer:

x + y - 5z = 52

Step-by-step explanation:

Given data:

Point on the plane (3,4,-9)

parallel to x + y -5z = 1

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the normal vector is (1 , 1 , -5)

formula for finding equation the plane given the normal vector (a,b,c) and point (m,n,o) is as

a(x - m) + b(y - n) + c(z-o)=0

Substituting the data we have

1(x-3) + 1( y- 4) - 5(z+9) =0

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3 years ago

12a^3b^2 +18a²b^2 – 12ab^2 Factor completely

AleksAgata [21]

The factorization of 12a^3b^2 +18a²b^2 – 12ab^2 is $6 a b^{2}(a+2)(2 a-1)$

Solution:

Given, expression is $12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}$

We have to factorize the given expression completely.

Now, take the expression

$12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}$

Taking $b^2$ as common term,

$b^{2}\left(12 a^{3}+18 a^{2}-12 a\right)$

Taking "a" as common term,

$b^{2}\left(a\left(12 a^{2}+18 a-12\right)\right)$

Taking "6" as common term,

$b^{2}\left(a\left(6\left(2 a^{2}+3 a-2\right)\right)\right)$

Splitting "3a" as "4a - a" we get,

$b^{2}\left(a\left(6\left(2 a^{2}+4 a-a-2\right)\right)\right)$

$\begin{array}{l}{b^{2}(a(6(2 a(a+2)-1(a+2))))} \\\\ {b^{2}(a(6((a+2) \times(2 a-1))))} \\\\ {6 a b^{2}(a+2)(2 a-1)}\end{array}$

Hence, the factored form of given expression is $6 a b^{2}(a+2)(2 a-1)$

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Step-by-step explanation:

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