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3 years ago

What's the least common denominator of 3⁄4, 4⁄5, and 2⁄3?

2 answers:

6 0

Because 5 is in this set, you can already exclude B. 2's factors can not be odd, so that excludes D. and 3's factors doesn't include 20, so you can exclude A.

Answer would be C.

Cloud [144]3 years ago

6 0

Answer:

Option (c) is correct.

least common common denominator is 60.

Step-by-step explanation:

Given numbers are $\frac{3}{4} \ ,\ \frac{4}{5} \ , \frac{2}{3}$

We have to find the least common denominator for the given numbers.

least common number is the greatest value that is multiple of all the given numbers.

Since, denominator of given fractional numbers are 3, ,4 ,5

So least common denominator is the number that is multiple of all the given denominator.

Since, 3 = 3 × 1

4 = 2 × 2

5 = 5 × 1

Least common number is find by multiplying all the factors and taking common factor one time.Since, 3, 4, 5 do not have any factor in common

So, the product of 3, 4, 5 are the least common denominator

thus, 3 × 4 × 5 = 60

Thus, least common common denominator is 60.

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ivanzaharov [21]

Answer:

The values of $x$ so that $y = \csc x$ have vertical asymptotes are $-2\pi$ , $-\pi$ , $0$ , $\pi$ , $2\pi$ .

Step-by-step explanation:

The function cosecant is the reciprocal of the function sine and vertical asymptotes are located at values of $x$ so that function cosecant becomes undefined, that is, when function sine is zero, whose periodicity is $\pi$ . Then, the vertical asymptotes associated with function cosecant are located in the values of $x$ of the form:

$x = 0\pm \pi\cdot i$ , $\forall \,i\in \mathbb{N}_{O}$

In other words, the values of $x$ so that $y = \csc x$ have vertical asymptotes are $-2\pi$ , $-\pi$ , $0$ , $\pi$ , $2\pi$ .

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3 years ago

For his birthday, Tyrone's parents gave him $7,790.00 which they put into a savings account that earns 15% interest compounded m

torisob [31]

Answer:

5 years and 5 months

Step-by-step explanation:

<u />

<u>Compound Interest Formula</u>

$\large \text{$ \sf A=P(1+\frac{r}{n})^{nt} $}$

where:

A = final amount
P = principal amount
r = interest rate (in decimal form)
n = number of times interest applied per time period
t = number of time periods elapsed

Given:

A = $17,474.00
P = $7,790.00
r = 15% = 0.15
n = 12
t = number of years

Substitute the given values into the formula and solve for t:

$\implies \sf 17474=7790\left(1+\dfrac{0.15}{12}\right)^{12t}$

$\implies \sf \dfrac{17474}{7790}=\left(1.0125}\right)^{12t}$

$\implies \sf \ln\left(\dfrac{17474}{7790}\right)=\ln \left(1.0125}\right)^{12t}$

$\implies \sf \ln\left(\dfrac{17474}{7790}\right)=12t \ln \left(1.0125}\right)$

$\implies \sf t=\dfrac{\ln\left(\frac{17474}{7790}\right)}{12 \ln \left(1.0125}\right)}$

$\implies \sf t=5.419413037...\:years$

Therefore, the money was in the account for 5 years and 5 months (to the nearest month).

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1 year ago

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atroni [7]

Answer:

Step-by-step explanation:

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