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wel
3 years ago
7

The sum of three consecutive multiples"of 8 is 192. Find the multiples.​

Mathematics
2 answers:
goblinko [34]3 years ago
8 0

Answer:

7,8,9

Step-by-step explanation:

192/8=24

X+(x+1)+(x+2)=24

3x+3=24

3x=21

x=7

so it’s 7,8,9

nika2105 [10]3 years ago
3 0

the more recent question you asked steps

Apply the fractions formula for multiplication, to

13×45

and solve

1×43×5

=415

Therefore:

13×45=415

Step-by-step explanation:

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Lelu [443]
Try this:  (29.99 units/day)(1000 days) = 29990 units.

Could you possibly include the given units of measurement in this post?

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3 years ago
6+5(a-1)=36 please answer me ​
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6 0
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3.59

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4 years ago
PLEASE HELP!! I WILL MARK THE FIRST CORRECT ANSWER BRAINLIEST!!))What is the volume of this cylinder?
Aleksandr-060686 [28]

Answer:

V = 6079.04 ft³

Step-by-step explanation:

The volume (V) of a cylinder is calculated as

V = πr²h ( r is the radius and h the height )

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V = 3.14 × 11² × 16

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6 0
2 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
4 years ago
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