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2 years ago

Diya used modeling to find

Mathematics

1 answer:

slamgirl [31]2 years ago

8 0

Answer:C

Step-by-step explanation: The quotient is correct because 27 tenths divided by 3 is 0.9 and 12 hundredths divided by 3 is 0.04

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90 POINTS!!!!! Given: KLIJ inscr. in k(O),m∠K = 64°, measure of arc LI = 69°, measure of arc IJ = 59°, measure of arc KJ =97°

Kay [80]

In any cyclic quadrilateral, angles opposite one another are supplementary, meaning

$m\angle K+m\angle I=m\angle L+m\angle J=180^\circ$

and given that $\boxed{m\angle K=64^\circ}$ , we have $\boxed{m\angle I=116^\circ}$ .

By the inscribed angle theorem,

$m\angle JLK=\dfrac12m\widehat{KJ}$

$m\angle ILJ=\dfrac12m\widehat{IJ}$

and since

$m\angle L=m\angle JLK+m\angle ILJ$

we have

$m\angle L=\dfrac{97^\circ+59^\circ}2\implies\boxed{m\angle L=78^\circ}$

and it follows that

$m\angle J=180^\circ-m\angle L\implies\boxed{m\angle J=102^\circ}$

5 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.

8 0

3 years ago

The product of -16 and an unknown number equals -256. What is the value of the unknown number?

nikitadnepr [17]

16 is the unknown value.

6 0

3 years ago

Evaluate: b ÷ a for a = 2 and b = 18

Jlenok [28]

Answer:

b/a=9, 9 is the answere

Step-by-step explanation:

i did it mentally lol

7 0

3 years ago

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Find the area of the shape in the photo.

Rashid [163]

First, you would start with finding the center of the trapezoid. You would multiply 1.6 by 2.4, leaving you with 3.84 sq ft for the center. Then, you subtract 1.6 from 3.2 because you want to find the area of the triangles. You would get 1.6. That means there is a triangle with a base of 1.6 feet and a height of 2.4 feet. The formula for calculating the area of a triangle is base times height divided by 2, and in this case it is 1.6 times 2.4 divided by 2. You would get 1.92, and to find the total amount of ft sq, you would add 1.92 and 3.84 to get 5.76 feet squared(don't forget to label!).

4 0

3 years ago

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