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2 years ago

.Sally is comparing 2 types of house insurance. Both companies provide the same service; however,

Mathematics

1 answer:

Nostrana [21]2 years ago

6 0

Answer:

car help

Step-by-step explanation:

In order to determine which type of house insurance would be considered as cheapest first it is necessary to determine either monthly or either weekly

As we know that

Generally in the month there is a 4 weeks

So For Car Help, the monthly would be

= $8.60 × 4

= $34.40

And for Car safety, the quoted price is $36.10 per week

So according to the above calculation the car help would be choose as the cheapest

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Find surface area of the prism

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SA=(a+b+c) h+bh
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=80+60
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2 years ago

What is the slope of the line

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(1, -3). (3, 0)

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Vanessa deposited money into a bank account that earned 1.25% simple interest each year. After 1/2 year, she had earned $5.00 in

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3 years ago

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Samuel drove 84 miles in 6 hours. How far can he drive in 6 hours?

rosijanka [135]

Answer:

the answer is 84 being how you asked the question but I dont think thats what you meant to ask. if thars wrong just tell me and ill answer it the way you need it

Step-by-step explanation:

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2 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

2 years ago