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3 years ago

Number 4 (Which proportion represents the problem below)

Mathematics

1 answer:

ololo11 [35]3 years ago

5 0

Answer:

$\textbf{D.}\quad\dfrac{43}{n}=\dfrac{17}{100}$

Step-by-step explanation:

You can start by translating the English statement to a math sentence:

17% × n = 43 . . . . . . 17% of a number is 43

To form a proportion like the answer choices, you can divide by n:

17/100 = 43/n . . . . . . matches selection D

_____

Of course, you know that 17% is the same as 17/100. (% is a shorthand way to write /100.)

_____

As a check on your answer, you can use your number sense to compare the size of n and 43. 17/100 = 43/n means that n will be larger than 43. 17/100 = n/43 means that 43 is larger than n. If 43 is a small fraction of n, then n must be larger than 43, so n/43 = 17/100 cannot be the appropriate proportion.

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A lifeguard is in a look out chair and sees a person in distress. The eye level of the lifeguard is 15 feet above the ground. Th

faltersainse [42]

Answer:

Horizontal distance between lifeguard and the person is 22 feet.

Step-by-step explanation:

Given: A lifeguard sees a person in distress.The eye level of the lifeguard is 15 feet above the ground. Angle of depression is 34°.

To find: Horizontal distance between lifeguard and the person.

Solution : If we draw a triangle then tan∅ = $\frac{height}{base}$

Here base is the horizontal distance.

Now we put the values in the formula.

tan 34° = $\frac{15}{base}$

Or Base = $\frac{15}{tan34}$

= $\frac{15}{0.675} = 22.22 feet$

So the answer is 22 feet.

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2 years ago

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3 years ago

A human heart beats an average of 80 beats per minute. How many heart beats is this in a day

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Answer:

115200

Step-by-step explanation:

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2 years ago

What is the mean for the data set?

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Read 2 more answers

CALC- limits 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

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

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3 years ago