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tigry1 [53]
3 years ago
9

Simplify exponents 5^-4over5^3 A.5^7 B.5^-1 C1/5 1/5^7

Mathematics
1 answer:
Dimas [21]3 years ago
5 0

Answer:

D. 1/5^(-7)

Step-by-step explanation:

ACCORDING TO THE LAW OF INDICES:

\frac{a {}^{n}  }{a {}^{m}  }  =  {a}^{n - m}

So; 5^(-4)/5^3

=5^(-4)-3

=5^-7

OR,

1/5^7

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baherus [9]

{4}^{5}   \div  {4}^{2}  \\  = {4}^{5 - 2}  \\  = 4^{3}

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2 years ago
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Consider the functions below.<br> f(x) = x² - 6x - 27<br> g(x) = x - 9
aleksklad [387]
Correct answer: 1) (f/g)(x)

Solution: f(x)=x^2-6x-27=(x+3)(x-9)

3 0
2 years ago
Find the area of the circle. leave your answer in terms is pi.
PilotLPTM [1.2K]

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9

Step-by-step explanation:

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3 years ago
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Find a nonzero vector orthogonal to the plane through the points: ????=(0,0,1), ????=(−2,3,4), ????=(−2,2,0).
ser-zykov [4K]

Answer:

The nonzero vector orthogonal to the plane is <-9,-8,2>.

Step-by-step explanation:

Consider the given points are P=(0,0,1), Q=(−2,3,4), R=(−2,2,0).

\overrightarrow {PQ}==

\overrightarrow {PR}==

The nonzero vector orthogonal to the plane through the points P,Q, and R is

\overrightarrow n=\overrightarrow {PQ}\times \overrightarrow {PR}

\overrightarrow n=\det \begin{pmatrix}i&j&k\\ \:\:\:\:\:-2&3&3\\ \:\:\:\:\:-2&2&-1\end{pmatrix}

Expand along row 1.

\overrightarrow n=i\det \begin{pmatrix}3&3\\ 2&-1\end{pmatrix}-j\det \begin{pmatrix}-2&3\\ -2&-1\end{pmatrix}+k\det \begin{pmatrix}-2&3\\ -2&2\end{pmatrix}

\overrightarrow n=i(-9)-j(8)+k(2)

\overrightarrow n=-9i-8j+2k

\overrightarrow n=

Therefore, the nonzero vector orthogonal to the plane is <-9,-8,2>.

8 0
3 years ago
I have a bottle of 3 l and a bottle of 5 l. How can I lose 7 l?
stiv31 [10]

Answer:

fill the 5l bottle up completely

pour the contents of the 5l bottle into the 3l until completely full

this should leave you with 2 litres in the 5 l bottle

empty out the water in the 3 litre bottle and then pour in the 2 litres

fill the 5 litre up completely and you have 7 litres

Step-by-step explanation:

3 0
2 years ago
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