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2 years ago

(x-5)(-2) Please help, need awnser by thursday

Mathematics

1 answer:

Amiraneli [1.4K]2 years ago

5 0

Answer:

−2+10

Step-by-step explanation:(

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Jaime makes the following claim.

Sphinxa [80]

Jaime is incorrect, the angle does not depend on the radius of the circles.

<h3>Is Jaime correct?</h3>

Remember that an angle that defines an arc on a circle, does not depend on the radius of the circle.

So, if we have an angle with a measure of π/3 radians in a circle with a radius of 3 inches and an angle with a measure of π/3 radians in a circle with a radius of 6 inches, these two angles are exactly the same thing.

The radius of the circle only has an impact on the length of the arc defined by the angle.

So Jaime is clearly incorrect.

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7 0

2 years ago

The mean spent on lottery tickets is normally distributed with a mean of $6.50 and a standard deviation of $2.25. If one custome

Natasha_Volkova [10]

First, we have to find the z scores of $4 and $9.50.
Z₁ = ($4 - $6.50)/$2.25 = -1.11
Z₂ = ($9.50 - $6.50)/$2.25 = 1.33

Then, using a z score table, we find the probability of 1.33 and -1.11, and subtract them to determine the probability in between.
0.9082 - 0.1335 = 0.7747 or 77.47%.

7 0

3 years ago

ANSWER FAST PLZ 33 POINTS!!!!!!!!!!!!!!!!!!

vlada-n [284]

Answer:

it is 706.5 in sq

Step-by-step explanation:

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3 years ago

Read 2 more answers

HELP ME PUT THESE IN ORDER TO LEAST TO GREATEST PLS ( i will give breanliest) HELP PLS

gregori [183]

Answer: $\sqrt{40}$ , $7\frac{1}{5}$ , $3\sqrt{6}$ , 15/2, 7.5

Step-by-step explanation: This is a bit tricky because 15/2 is equal to 7.5 !!

√40 = 6.32455532

7 1/5 = 7.2

3√6 = 7.348469228

15/2 = 7.5

7.5 = 7.5

6 0

3 years ago

Read 2 more answers

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 <em>too easy</em> 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.

<u>Differential</u> (Power Rules)

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

<u>Differentiation</u> (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)}}$

<u>Differential</u> (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.

8 0

3 years ago