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3 years ago

Read the passage.

Mathematics

1 answer:

alexandr1967 [171]3 years ago

6 0

Answer:

D sentence 5

Step-by-step explanation:

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Write in terms of i. Simplify your answer as much as possible. ✓-99

telo118 [61]

The difference is 99 percent off and

6 0

3 years ago

$3000 is invested at 3% annual interest for 3 years. a. Simple Interest b. Compound Interest

stich3 [128]

Answer:

a. $270

b. $3,278.18

Step-by-step explanation:

Given that

The principal amount is $3,000

Annual rate of interest is 3%

And, the time period is 3 years

We need to find out the simple interest & compound interest

The following formulas should be used

a. For simple interest

= Principal × rate of interest × time period

= $3,000 × 3% × 3 years

= $270

b. For compound interest

= Principal × (1 + rate of interest)^time period

= $3,000 × (1 + 0.03)3

= $3,000 × 1.03^3

= $3,278.18

8 0

2 years ago

Find the prime factorization of each number: 11. 29 12. 168

Ray Of Light [21]

29 is a prime number.
Factors of 29: 1 , 29

168 is a composite number, therefore you can find the prime factorization for it. Divide prime factors until what is left is a prime number. Use the tree factor. See attached picture:

Prime factorization of 168: 2, 2, 2, 3, 7

7 0

3 years ago

What are 2 factors of 72y + 108z?

emmasim [6.3K]

72y+108z Factored is 36(2y+3z). So I guess the 2 factors will be 2y and 3z??

7 0

3 years ago

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.

grandymaker [24]

$(2x+1)^{\cot x}=\exp\left(\ln(2x+1)^{\cot x}\right)=\exp\left(\cot x\ln(2x+1)\right)=\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)$

where $\exp(x)\equiv e^x$ .

By continuity of $e^x$ , you have

$\displaystyle\lim_{x\to0^+}\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)$

As $x\to0^+$ in the numerator, you approach $\ln1=0$ ; in the denominator, you approach $\tan0=0$ . So you have an indeterminate form $\dfrac00$ . Provided the limit indeed exists, L'Hopital's rule can be used.

$\displaystyle\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\frac2{2x+1}}{\sec^2x}\right)$

Now the numerator approaches $\dfrac21=2$ , while the denominator approaches $\sec^20=1$ , suggesting the limit above is 2. This means

$\displaystyle\lim_{x\to0^+}(2x+1)^{\cot x}=\exp(2)=e^2$

7 0

2 years ago