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3 years ago

HELP I NEED HELP ASAP

Mathematics

1 answer:

algol [13]3 years ago

5 0

Answer:

why do people always put links

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Find the balance in the account after the given period.

leva [86]

Substituting our values into the formula we get
$A=4200(1+ \frac{0.042}{12})^1^2^*^2
\\ = 4200(1+0.0035)^2^4
\\ = 4200(1.0035)^2^4
\\ = 4567.37$

7 0

3 years ago

(-7a^4bc^3)(5ab^4c^2)

taurus [48]

-35a^5b^5c^5 is the answer

6 0

3 years ago

Can someone pls pls pls pls pls pls answer these question

Lisa [10]

Answer:

Decay

Step-by-step explanation:

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3 years ago

Simplify the expression using the exponent properties 5 ^ −4 x 5^ −3

vodka [1.7K]

Answer:

$5^{-7}$ or 0.0000128

Step-by-step explanation:

Add the exponents together

-4+-3=-7

3 0

4 years ago

HELP PLEASE !!!!!!!!!!!

horrorfan [7]

Answer:

option B

Step-by-step explanation:

$\lim_{x\to \infty} f(x) = \lim_{x \to \infty} (log(x+2)-3 )$

$= \lim_{x \to \infty} ( log(x+2)) - \lim_{x \to \infty} 3\\\\=\infty - 3\\\\= \infty$

$\lim_{x \to 2} f(x) = \lim_{x \to 2} ( log(x+2)) - \lim_{x \to2} 3\\\\$

$= log( 2 + 2 ) - 3\\\\= log 4 - 3\\\\=log(2^2) -3\\\\=2 log 2 -3$

$\lim_{x \to -2} f(x) = \lim_{x \to -2} ( log(x+2)) - \lim_{x \to -2} 3\\\\$ $\lim_{x \to -2}_+ f(x) = \lim_{x \to -2_+} ( log(x+2)) - \lim_{x \to -2_+} 3 = -\infty\\\\\lim_{x \to -2_-} f(x) = \lim_{x \to -2}_- ( log(x+2)) - \lim_{x \to -2}_- 3 = -\infty\\\\\\\lim_{x \to -2} f(x) = - \infty$

6 0

3 years ago