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2 years ago

Why does the fraction 20/25 not belong with 1/20, 2/3 and 5/4?

Mathematics

1 answer:

devlian [24]2 years ago

7 0

I’m not 100% sure but I think it’s because 20/25 can still be simplified further, while the other 3 can’t.

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How do I express 540 as a product of prime factors

aksik [14]

540=
54*10=
6*9*2*5=
2*3*3*3*2*5=
2*2*3*3*3*5 or in exponential form
(2²)(3³)(5)

6 0

3 years ago

Read 2 more answers

Jimmy earns $9.25 per hour working at a local grocery store. If Jimmy earned $74 last week, which equation would be used to dete

Alekssandra [29.7K]

9.25h=74 you multiply h by 9.25

8 0

3 years ago

An artist wants to frame a square painting with an area of 400 square inches . she wants

Artemon [7]

(a) $x^{2}$ = 400
(b) $\sqrt{x^2} = \sqrt{400}$
$x=20,-20$
(c) Only the positive solution makes sense because you cannot have a negative side length.
(d) Since one side of the painting is 20 inches, all four sides together will be 20*4 = 80 inches

6 0

3 years ago

Consider this linear function:

ivanzaharov [21]

Answer:

(-8,-3)

(-4,-1)

(0,1)

(2,2)

(6,4)

Step-by-step explanation:

One by one, substitue the x values and solve for y

3 0

2 years ago

If lim x-> infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b

MAXImum [283]

We have

$\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}$

$\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0$

The rational term vanishes as x gets arbitrarily large, so we can ignore that term, leaving us with

$\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0$

and this happens if a = 1 and b = -1.

To confirm, we have

$\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0$

as required.

3 0

2 years ago