2 years ago

A^8/a^3 hey can someone help me

Mathematics

1 answer:

Stells [14]2 years ago

3 0

A^8/a^3=a^5 you subtract the exponents

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What is the coeffieent of (x^4)(y^4) in the expansion of (x+y)^8

kap26 [50]

ANSWER

The coefficient is 70.

EXPLANATION

The given binomial expression is

$(x+y)^8$

The specific term of a binomial expansion can be determined using the formula,

$T_{r+1}= nC_ra^{n-r}b^r$
where

$n=8,r=4,a=x,b=y$

We substitute these values into the formula to get,

$T_{4+1}= 8C_4x^{8-4}y^4$

We simplify to get,

$T_{5}= 70x^{4}y^4$

Hence the coefficient is 70.

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2 years ago

What are the solutions to the following equation?

gregori [183]

Answer: -8.5

Step-by-step explanation:

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2 years ago

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Simplifying expressions:<br> Simplify 22-5(2x+4)

tangare [24]

22-10x-20

=2-10x

=2(1-5x)

hope this helps.

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2 years ago

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Given $a \equiv 1 \pmod{7}$, $b \equiv 2 \pmod{7}$, and $c \equiv 6 \pmod{7}$, what is the remainder when $a^{81} b^{91} c^{27}$

Blizzard [7]

$\begin{cases}a\equiv1\pmod7\\b\equiv2\pmod7\\c\equiv6\pmod7\end{cases}$

$a^{81}\equiv1^{81}\equiv1\pmod7$

$b^{91}\equiv2^{91}\pmod7$

$2^{91}\equiv(2^3)^{30}\times2^1\equiv8^{30}\times2\pmod7$
$8\equiv1\pmod7$
$2\equiv2\pmod7$
$\implies2^{91}\equiv1^{30}\times2\equiv2\pmod7$

$c^{27}\equiv6^{27}\pmod7$

$6^{27}\equiv(-1)^{27}\equiv-1\equiv6\pmod7$

$\implies a^{81}b^{91}c^{27}\equiv1\times2\times6\equiv12\equiv5\pmod7$

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2 years ago

Find the value of x in the given

iren2701 [21]

Answer:

You divide 62/12=5.166

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2 years ago