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3 years ago

Whats the correct answer ?

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

3 0

Answer:

$80, $48

Step-by-step explanation:

The original price was x.

The price is reduced by 40%. The markdown is 40% of x, or 0.4x.

The actual dollar amount of the markdown is $32.

That means that 0.4x must equal $32.

We set up an equation and solve for x, the original price.

0.4x = 32

Divide both sides by 0.4

x = 32/0.4

x = 80

The original price is $80.

The markdown is $32.

$80 - $32 = $48

The marked down price is $48.

Answer: $80, $48

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4 times what equals 204? Explanation please.

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To find the answer to your question you just have to divide 204 by 4:

204 / 4 = 51

It means that 4 times 51 equals 204:

4 * 51 = 204

I hope it will help you :)

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2 years ago

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Write the equation of the line passing through (0,-8),with slope of -1/2, in slope-intercept form _____?

Vadim26 [7]

The equation would be Y= 1/2x -8

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2 years ago

I need help on this

Naily [24]

Answer:x=90° y=58° z=32°

Step-by-step explanation:

x=90

y=180-(90+32) sum of angles in a triangle is 180

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3 years ago

Two sides of a triangle measure 14 inches and 8 inches. which cannot be the length of the remaining side?

horrorfan [7]

1. The "triangle inequality" states that any of the sides of the triangle, is larger than the difference of the other 2 sides (the larger minus the smaller), and it is smaller than the sum of them.

2. So the third side of this triangle must be larger than 14-8=6 (in) and smaller than 14+8=22 (in)

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2 years ago

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:

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So our limit becomes

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2 years ago