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3 years ago

Need help ASAP thanks !!! 20 points !!

Mathematics

2 answers:

iVinArrow [24]3 years ago

6 0

Answer:

Step-by-step explanation:

PQ = 3

P'Q' = 4

Answer: B.

Sati [7]3 years ago

3 0

idk im not sure but i think its b

You might be interested in

PLEASE HELP!!! WILL GIVE BRANLIEST

Rus_ich [418]

Check the picture below, so the hyperbola looks more or less like that.

based on the provided vertices, its center is at the origin, as you see there, the "a" component or traverse axis is 4, the "c" distance is 5.

$\bf \textit{hyperbolas, vertical traverse axis }
\\\\
\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h, k\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2 + b ^2}
\end{cases}\\\\
-------------------------------$

$\bf \begin{cases}
h=0\\
k=0\\
c=5\\
a=4
\end{cases}\implies \cfrac{(y-0)^2}{4^2}-\cfrac{(x-0)^2}{b^2}=1
\\\\\\
\stackrel{c}{5}=\sqrt{4^2+b^2}\implies 5^2=16+b^2\implies 25=16+b^2
\\\\\\
9=b^2\implies \sqrt{9}=b\implies \boxed{3=b}
\\\\\\
\cfrac{(y-0)^2}{4^2}-\cfrac{(x-0)^2}{3^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{9}=1$

3 0

4 years ago

Read 2 more answers

A={ 0,1,3,5,6}; B={1,2,3,7} draw the venn diagram of AUB Writ

Sati [7]

Answer:

Here,

n(A)={0,1,3,5,6}

n(B)={1,2,3,7}

n(AUB)=?

we know that,

AUB={0,1,2,3,5,6,7}

Step-by-step explanation:

is it okay ?

hopefully it works <3 ❣️

8 0

2 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

Round the answer using proper significant figures. <img src="https://tex.z-dn.net/?f=%5Cfrac%7B15.58701%7D%7B2.638%7D" id="T

Maslowich

Answer:

5.909

Step-by-step explanation:

The quotient is approximately ...

15.58701/2.638 ≈ 5.9086467...

The least-precise contributor to this calculation has 4 significant digits, so it would be appropriate to round the result to 4 significant digits:

5.909

_____

Additional comment

The quotient has a 659-digit repeating decimal fraction starting with 6467....

7 0

3 years ago

3. PPPPPPPPPPPPPLLLLLLLLLLLLLLZZZZZZZZZZZZZZZ HELP!!!!!!!!! A. B. C. D.

Vladimir [108]

Answer:

a............................................................................................................

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers