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Ymorist [56]
2 years ago
14

PLESE HELP ASAP!!!! Worth 15 point plz help!!

Mathematics
2 answers:
yulyashka [42]2 years ago
8 0
The last expression is correct
Y_Kistochka [10]2 years ago
3 0

Answer:

The last expression is correct

Step-by-step explanation:

We have been given the following expression for simplification;

4^{4}*4^{-9}

The bases of the two expressions are the same. They are both equal to 4. When multiplying numbers with the same base, we simply add their exponents. In this case our exponents are;

4 and -9

adding them we have;

4 + (-9) = -5

The simplified expression thus becomes;

4^{-5}=\frac{1}{4^{5}}

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Find 2 numbers a and b such that (a-b)^2 < (a+b)(a-b)< (a+b)^2 Find two numbes a and b such that this is not true
Svet_ta [14]

Suppose for the moment that the inequality holds for all a,b:

(a-b)^2

Expanding everything gives

a^2-2ab+b^2

\implies-ab

In particular, the inequality says that -ab for any choice of a,b. But if a and b>0, then -ab>0 while ab.

So one possible choice of a,b could be a=-1 and b=1. Then we get

(-1-1)^2=4

(-1+1)(-1-1)=2

(-1+1)^2=0

but clearly it's not true that 4.

8 0
3 years ago
Consider three events A, B and C with following properties.
lisov135 [29]
By the inclusion/exclusion principle,

\mathbb P(D)=\mathbb P(A\cup B\cup C)
\mathbb P(D)=\mathbb P(A)+\mathbb P(B)+\mathbb P(C)-\bigg(\mathbb P(A\cap B)+\mathbb P(A\cap C)+\mathbb P(B\cap C)\bigg)+\mathbb P(A\cap B\cap C)
\mathbb P(D)=\dfrac14+\dfrac16+\dfrac14-\dfrac39+\dfrac1{13}
\mathbb P(D)=\dfrac{16}{39}\neq1

So the union of A, B, and C does not constitute the entire sample space.
5 0
3 years ago
How to write 20/10 as a mixed number
Rasek [7]
You can't the answer is 2 because 20 goes into 10 twice
8 0
3 years ago
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The number of bacteria in a certain population increases according to a continuous exponential
ludmilkaskok [199]

Answer:

Step-by-step explanation:

4 0
2 years ago
Please help, I have been trying for a while :(<br><br>question on photo​
Zigmanuir [339]

Answer:

\frac{x-22}{(x+2)(x-4)}

Step-by-step explanation:

multiply the numerator/denominator of the first fraction by (x - 4)

multiply the numerator/denominator of the second fraction by (x + 2)

this ensures that the fractions have a common denominator

\frac{4}{x+2} - \frac{3}{x-4}

= \frac{4(x-4)}{(x+2)(x-4)} - \frac{3(x+2)}{(x+2)(x-4)} ← subtract numerators leaving the common denominator

= \frac{4x-16-3x-6}{(x+2)(x-4)}

=\frac{x-22}{(x+2)(x-4)}

4 0
2 years ago
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