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1 year ago

If x-2y=10 and x≠0, what is the value of 2x/y+5

Mathematics

1 answer:

VladimirAG [237]1 year ago

5 0

Answer: 4

Step-by-step explanation:

x - 2y =10

x = 10 + 2y

$\frac{2x)}{y + 5}$

$\frac{2(10+2y)}{y+5}$

$\frac{20+4y}{y+5}$

$\frac{4(5+y)}{y+5}$ ... cancel like terms (y + 5)

= 4

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Two more than 5 times a number is greater than or equal to 3

jenyasd209 [6]

Answer:

n ≥ 1/5

Step-by-step explanation:

Let the number be n. Then 5n + 2 ≥ 3.

If you want to solve this for n, then subtract 2 from both sides, obtaining:

5n ≥ 1. Isolate n by dividing both sides by 5. Then we have n ≥ 1/5

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3 years ago

0. Kevin needs to buy shrubs for his yard. Six shrubs at the nursery cost $72.00. At the garden

Elanso [62]

Answer:

72÷6=12

so at the nursery it cost $12 per shrub

66÷6=11

so at the store it cost $11 per shrub

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but if you buy 21 shrubs from the store it only cost: 11×21=$231

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2 years ago

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What value of x would make the figure a parallelogram (show your math please)

Verdich [7]

Answer:

Step-by-step explanation:

Diagonals of a parallelogram bisects each other.

Therefore,

4x = 8

x = 8/4

x = 2

5x = 3x + 4

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2x = 4

x = 4/2

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2 years ago

What is the answer to 1/4N=24

PilotLPTM [1.2K]

0.25*N=24
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3 years ago

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CALC- limits<br> please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

3 years ago