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yulyashka [42]
3 years ago
11

write the number for each named while number. fifty-two trillion one billion,nineteen million,five hundred two thousand,eleven

Mathematics
1 answer:
tatuchka [14]3 years ago
3 0

52,001,019,502,011. I hope this helps!

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PLEASE ANSWER ASAP FOR BRAINLEST!!!!!!!!!!!!!!!!1
posledela

<u>Assuming that k is variable:</u>

6k - 6²

<u>Assuming that 3k represents 3000:</u>

426×7

7 0
3 years ago
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Philip save $8 each month. how many month will it take him to save at least $60
Vlada [557]

Answer:

About 8 months

Step-by-step explanation:

$60 divided by $8 is 7.5.

Round it up to 8 since you cannot have 7.5 months. Do not round it down because 7 x 8 is 56. Hope this helped.

8 0
3 years ago
CAN SOMEONE PLEASE HELP, ILL GIVE BRAINLIEST.
ratelena [41]

Answer:

Line RT || Line VS ∠SRV ≅ ∠TUR - Given/Alternate Interior Angles Theorem

∠TRU ≅ ∠SVR - Corresponding Angles Theorem

ΔRTU ~ ΔVSR - AA Similarity Theorem

Step-by-step explanation:

The statement Line RT || Line VS ∠SRV ≅ ∠TUR is given. This can be explained with the Alternate Interior Angles Theorem. It states that if two parallel lines (TR and VS) are cut by a transversal (∠SRV), then the pairs of alternate interior angles are congruent.

∠TRU and ∠SVR correspond to each other, so that would be the Corresponding Angles Theorem.

That leaves ΔRTU ~ ΔVSR being that away due to the AA Similarity Theorem. It states that states that if two angles of one triangle (∠RTU, ∠URT for example) are congruent to two angles of another triangle (∠VSR, ∠RVS are the congruent angles to the two from before), then the triangles are similar.

5 0
3 years ago
How do you solve this limit of a function math problem? ​
hram777 [196]

If you know that

e=\displaystyle\lim_{x\to\pm\infty}\left(1+\frac1x\right)^x

then it's possible to rewrite the given limit so that it resembles the one above. Then the limit itself would be some expression involving e.

For starters, we have

\dfrac{3x-1}{3x+3}=\dfrac{3x+3-4}{3x+3}=1-\dfrac4{3x+3}=1-\dfrac1{\frac34(x+1)}

Let y=\dfrac34(x+1). Then as x\to\infty, we also have y\to\infty, and

2x-1=2\left(\dfrac43y-1\right)=\dfrac83y-2

So in terms of y, the limit is equivalent to

\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^{\frac83y-2}

Now use some of the properties of limits: the above is the same as

\displaystyle\left(\lim_{y\to\infty}\left(1-\frac1y\right)^{-2}\right)\left(\lim_{y\to\infty}\left(1-\frac1y\right)^y\right)^{8/3}

The first limit is trivial; \dfrac1y\to0, so its value is 1. The second limit comes out to

\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^y=e^{-1}

To see why this is the case, replace y=-z, so that z\to-\infty as y\to\infty, and

\displaystyle\lim_{z\to-\infty}\left(1+\frac1z\right)^{-z}=\frac1{\lim\limits_{z\to-\infty}\left(1+\frac1z\right)^z}=\frac1e

Then the limit we're talking about has a value of

\left(e^{-1}\right)^{8/3}=\boxed{e^{-8/3}}

# # #

Another way to do this without knowing the definition of e as given above is to take apply exponentials and logarithms, but you need to know about L'Hopital's rule. In particular, write

\left(\dfrac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\ln\left(\frac{3x-1}{3x+3}\right)^{2x-1}\right)=\exp\left((2x-1)\ln\frac{3x-1}{3x+3}\right)

(where the notation means \exp(x)=e^x, just to get everything on one line).

Recall that

\displaystyle\lim_{x\to c}f(g(x))=f\left(\lim_{x\to c}g(x)\right)

if f is continuous at x=c. \exp(x) is continuous everywhere, so we have

\displaystyle\lim_{x\to\infty}\left(\frac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}\right)

For the remaining limit, write

\displaystyle\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}=\lim_{x\to\infty}\frac{\ln\frac{3x-1}{3x+3}}{\frac1{2x-1}}

Now as x\to\infty, both the numerator and denominator approach 0, so we can try L'Hopital's rule. If the limit exists, it's equal to

\displaystyle\lim_{x\to\infty}\frac{\frac{\mathrm d}{\mathrm dx}\left[\ln\frac{3x-1}{3x+3}\right]}{\frac{\mathrm d}{\mathrm dx}\left[\frac1{2x-1}\right]}=\lim_{x\to\infty}\frac{\frac4{(x+1)(3x-1)}}{-\frac2{(2x-1)^2}}=-2\lim_{x\to\infty}\frac{(2x-1)^2}{(x+1)(3x-1)}=-\frac83

and our original limit comes out to the same value as before, \exp\left(-\frac83\right)=\boxed{e^{-8/3}}.

3 0
3 years ago
The question is below
Tanzania [10]

Answer:

B. $4.75

Step-by-step explanation:

3(4) + 7 = 19

19/4=4.75

7 0
2 years ago
Read 2 more answers
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