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3 years ago

Four less then five times a number is equal to 11

Mathematics

1 answer:

choli [55]3 years ago

4 0

4 less (-4) 5 times a number (5n) is (=) 11 (11)

-4+5n=11
add 4 to both sides
4-4+5n=11+4
0+5n=15
5n=15
divide both sides by 5
5n/5=15/5
1n=3
n=3
the number is 3

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The first tern of an arithmetic sequence is -3 and the fifteenth term id 53. What is the common difference of the sequence?

arsen [322]

The common difference is 4. U know that the formula is the first term + d(n -1). N-1 is 14 so u know that 14 times a number needs to equal 56 because the first term is -3 and 56-3 is 53. 14 * 4 is 56. So the common difference is 4

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2 years ago

Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4

djverab [1.8K]

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

$\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty$

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

$\lim_{x \to \infty}\frac{x^3}{(x-2)^4}$

if $\lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b$ then y = b is horizontal asymptote

Then:

$\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0$

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

3 0

3 years ago

Read 2 more answers

Question 21 is probably the hardest question on this homework

Snezhnost [94]

-15+5x= 20

5x= 35
x= 7

20= -15+5x

35= 5x

7 = x

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2 years ago

Read 2 more answers

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I think it is A
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If you run a marathon of 26 1/5 miles in a time of 5 4/5 how far do you run in 1 hour<br><br>pls

alisha [4.7K]

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