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Elanso [62]
3 years ago
14

What is 9(x + 8) + x(5+4)

Mathematics
2 answers:
lina2011 [118]3 years ago
7 0

Answer:

72+ 18x

Step-by-step explanation:

9x+72+5x+4x

72+ 18x

Softa [21]3 years ago
7 0
9(x + 8) + x(5 + 4)

9x + 72 + 5x + 4x

18x + 72

(If you have been asked to solve for x then also include the following, if not then only write what is above!)

18x + 72 = 0

18x = -72

x = -4
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Sandra is making decorations for the school dance out of construction paper. Each decoration uses 223 2 2 3 sheets of paper. San
scoundrel [369]

Answer:

Step-by-step explanation:

CACA POOP

7 0
3 years ago
(help) Miller’s Farm had apples on sale for $1.78 a pound. Maria spent $15.21 on apples. How many pounds did she buy?
jolli1 [7]

Answer:

8.5 pounds

Step-by-step explanation:

$ 1.78 - 1 pound

$ 15.21 - 15.21/1.78

= 8.5

5 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
3 years ago
Ann and Tom want to establish a fund for their​ grandson's college education. What lump sum must they deposit at an 8.2​% annual
GaryK [48]
Compound interest formula = a=P(1+r/n)^nt

P= lump sum to deposit (solving for)

A= amount accumulated over the entire time (20000)

n= number of times interest is compounded annually (1)

r= rate of interest (0.82)

T= total number of years (15)

20000=P(1+0.082/1)^1*15

20000=P(1.082)^15
20000=P(3.26143638)
20000/3.26143638=P
P=$6132.2674
7 0
3 years ago
Read 2 more answers
-8)(56)=1<br><br> What is the missing value that makes the equation true?
Rudik [331]
the correct answer is -448
7 0
3 years ago
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