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3 years ago

PLEASE HELP!!!!

Mathematics

1 answer:

saveliy_v [14]3 years ago

8 0

Answer:

Hi there!

Your answers are:

1) 38^10

2) 20^3

Step-by-step explanation:

1) Theres a property that goes like this:

(x^m)^n = x^mn

When you do a power of a power, you multiply them together!

2) There's a property that looks like this:

x^ m / x^n = x^ m-n

When dividing exponents, subtract!

Hope this helps!

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What set of values is apart of the solution set to the inequality 3.5p+14>7p

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Answer:

Step-by-step explanation:

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3 years ago

I just need 30 please

PIT_PIT [208]

In the smaller triangle, let $y$ be the length of the smallest leg. Then the smallest leg in the larger triangle is $14-y$ .

Within the scope of the smaller triangle, we have $y$ such that

$x^2+y^2=13^2\implies y=\sqrt{13^2-x^2}$

Then within the larger the triangle, we would have

$x^2+(14-y)^2=15^2\iff x^2+\left(14-\sqrt{13^2-x^2}\right)^2=15^2$

Now we can solve for $x$ :

$x^2+14^2-28\sqrt{13^2-x^2}+(13^2-x^2)=15^2$

$\implies x=12$

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3 years ago

Order the numbers from least to greatest.<br><br> 0.70, 0.701, 0.71, 0.078

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Answer:

0.078,0.70,0.701,0.71

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3 years ago

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4 years ago

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A single card is drawn from a standard 52-card deck. Let D be the event that the card drawn is a black card and let F be the eve

tester [92]

Answer:

The indicated probability of $P(D \cup F')=\frac{25}{26}$

Step-by-step explanation:

Probability of an event E to be;

P(E) = $\frac{Number of events within E}{Total number of possible outcomes}$

As per the given condition:

Total number of possible outcomes = 52 cards.

Let the event be D and F as follows;

D : Drawn card is a black card

F : Drawn card is a 10 card.

Then,

From the given condition:

P(D) = $\frac{26}{52}$ [Out of 52 cards, 26 were black] ,

P(F) = $\frac{4}{52}$ [Out of 52 cards, there are four 10 cards]

For any two events A and B we always have;

$P(A \cup B) = P(A)+P(B)-P(A \cap B)$

Now, we have to find the indicated probability:

$P(D \cup F')=P(D)+P(F')-P(D \cap F')$ ......[1]

First find the P(F');

P(F') =1-P(F) = $1-\frac{4}{52} =\frac{52-4}{52} =\frac{48}{52}$

Also, to find $P(D \cap F')$ .

We use the formula :

For any event A and B independent variable.

$P(A \cap B) =P(A) \cdot P(B)$

then;

$P(D \cap F') =P(D) \cdot P(F') = \frac{26}{52} \cdot \frac{48}{52} =\frac{24}{52}$

Now, substitute these in [1];

$P(D \cup F')=\frac{26}{52} +\frac{48}{52} -\frac{24}{52}$ = $\frac{26+48-24}{52} =\frac{50}{52} = \frac{25}{26}$

Therefore, the probability of $P(D \cup F')=\frac{25}{26}$

3 0

3 years ago