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3 years ago

I need help finding out what AC is!!

Mathematics

2 answers:

zubka84 [21]3 years ago

7 0

Answer:

Step-by-step explanation:

Count the spaces between A and C. There are 5 of them.

There are 3 spaces between A=-3 and 0, and there are 2 more between 0 and C=2. Thus AC = 3+2 = 5.

In arithmetic terms, the distance is the difference ...

AC = C - A = 2 -(-3) = 2+3

AC = 5

iragen [17]3 years ago

5 0

I think it’s b.5. Hope this helps. :)

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33 POINTS! Right answer gets brainliest Find the area of the shape below.

Dvinal [7]

Answer: 112.5 m^2

Step-by-step explanation:

Area of rectangle = lw

l = 7.5

w = 15

7.5 * 15 = 112.5

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2 years ago

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PLEASE HELP ME!!!!!!!!

GREYUIT [131]

Answer:

The solution to the system of equations (x, y) = (2, 4) represents the month in which exports and imports were equal. Both were 4 in February.

Step-by-step explanation:

We're not sure what "system of equations" is being referenced here, since no equations are shown or described.

Perhaps your "system of equations" is ...

f(x) = some equation

g(x) = some other equation

Then the solution to this system of equation is the pair of values (x, y) that gives ...

y = f(x) = g(x)

If x represents the month number, then the solution can be read from the table:

(x, y) = (2, 4)

This is the month in which exports and imports were equal. Both numbers were 4 in February.

7 0

3 years ago

How do you find the limit?

coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s too easy but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

Differential (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

Differentiation (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

Differential (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$

Now we can substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}$

Thus, the limit value is 2/5 same as the first method.

Notes:

If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.

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2 years ago

For consumers making purchases online, 60% have devices made by Apple, 85% own a smartphone, and 75% use Venmo. Also, out of the

kompoz [17]

Answer:

The probability that a customer selected at random has an Apple device or own a smartphone or both is 0.94

Step-by-step explanation:

The percentage of costumers that have a device made by Apple = 60%

The percentage of customers that own a smartphone = 85%

The percentage of customers that use Venmo = 80%

The percentage out of the smartphone users that use Venmo = 80%

The probability both independent events A and B occurring = P(A) × P(B)

The exclusive probability of A or B occurring P(A XOR B) = P(A) + P(B) - 2 × P(A ∩ B)

Therefore, the probability of A or B or both occurring is given as follows;

P(A or B or Both) = P(A) + P(B) - 2 × P(A ∩ B) + P(A) × P(B)

Where A represent the percentage of costumers that have a device made by Apple and let B represent the percentage of users that have a smartphone, we have;

P(A or B or Both) = 0.6 + 0.85 - 2×0.6×0.85 + 0.6 × 0.85 = 0.94

Therefore, the probability that a customer selected at random has an Apple device or own a smartphone (or both), P(A or B or Both) = 0.94

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3 years ago

Help PLEASE if you know the correct answer PLease I give brainliest

il63 [147K]

Answer:

the answer is c mathematical way

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2 years ago

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