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emmasim [6.3K]
3 years ago
7

To solve 6x=72 do you use the addition property of equality

Mathematics
1 answer:
lawyer [7]3 years ago
3 0

Answer:

No , the division property of equality.

Step-by-step explanation:

The Division Property of Equality states that if you divide both sides of an equation by the same nonzero number, the sides remain equal.

Hope this helps!!

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50 POINTS PLEASE HELP Segment JK has endpoint J(2,4) and K(6,2). You dilate the segment using a dilation centered at the origin
stira [4]

Answer:

C.

Step-by-step explanation:

7 0
3 years ago
What is the expansion of (3+x)^4
Vlad1618 [11]

Answer:

\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81

Step-by-step explanation:

Considering the expression

\left(3+x\right)^4

Lets determine the expansion of the expression

\left(3+x\right)^4

\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i

a=3,\:\:b=x

=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i

Expanding summation

\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}

i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0

i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1

i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2

i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3

i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4

=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4

=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4

as

\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81

\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x

\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2

\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3

\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4

so equation becomes

=81+108x+54x^2+12x^3+x^4

=x^4+12x^3+54x^2+108x+81

Therefore,

  • \left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81
6 0
3 years ago
Select the correct answer.
Genrish500 [490]
B 16 miles an hour

Hope this helps
3 0
2 years ago
-6y+y2+3-4y2-7+1-4y<br> Please help
vaieri [72.5K]
-6y+y2+3-4y2-7+1-4y in simplified form is -10y-3y^2-3
5 0
2 years ago
Solve for x:<br> -14x &gt; -3
Alexus [3.1K]

Answer:

x>11

Step-by-step explanation:

-14x>-3

+14  +14

x > 11

6 0
3 years ago
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