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3 years ago

I need the answers for this assignment pls help :(

Mathematics

1 answer:

Kruka [31]3 years ago

3 0

Answer:

115

Step-by-step explanation:

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Mrs. Jacobs purchased a 5 pound package of ground beef for $12.40. She decided to use 8 ounces each day for dinner recipes. What

WITCHER [35]

There is 16 ounces in 1 pound. so you would want to divide 12.40 by 5 = $2.48 . so its $2.48 for a pound divide by 2 and you answer would be $1.24... i hope that helps

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3 years ago

A new car is purchased for 19900 dollars. The value of the car depreciates at 7.5% per year. What will be the value of the car a

bogdanovich [222]

Answer:

$10665,64

Step-by-step explanation:

S=19900*(1-(7,5/100))^8=10665, 64

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3 years ago

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Given F(x) = -x^2+10x-3, find f(9)

soldi70 [24.7K]

Answer:

x =(10-√112)/2=5-2√ 7 = -0.292

x =(10+√112)/2=5+2√ 7 = 10.292

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3 years ago

What is the work for 34,992 divided by 81

bagirrra123 [75]

34992/81 or 81*x=34992 = 432

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3 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:

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

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