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3 years ago

Ivanna has to carry 308 apples from a farm to the market. How many baskets will she need, given that each basket

Mathematics

1 answer:

Arisa [49]3 years ago

7 0

Answer: I would say 9 baskets

Step-by-step explanation: if you divide 308 by 36, it would be 8.55. Logically speaking, you can’t have 8.5 baskets. So 9 it is.

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B9%7D" id="TexFormula1" title="\frac{1}{9}" alt="\frac{1}{9}" align="absmiddle

Jobisdone [24]

$\dfrac{1}{9} - \dfrac{5}{6}$

Set up

$\dfrac{2}{18} - \dfrac{15}{18}$

Get all fractions to a common denominator which will allow you to subtract them

$\dfrac{2 - 15}{18}$

Subtract numerators

$\boxed{- \dfrac{13}{18}}$

Simplify

4 0

3 years ago

Please help. im completey lost

Radda [10]

Answer:

I had it before and it is 15 ft²

8 0

3 years ago

Read 2 more answers

What is the measure of ∠S? 50° 30° 100° 60°

Scilla [17]

Hello,

Answer B

An exterior angle of a circle has like measure the half of difference of the arcs.
mes S=(85°-25°)/2=60°/2=30°

7 0

3 years ago

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

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

3 0

3 years ago

A lamb is on sale, and it's price is reduced from $80 to $50, what is the percent decrease?

Readme [11.4K]

Answer:

.375% decrease also in fraction form 3/8

Step-by-step explanation:

so if 80 is 100 percent 3/8 of it was decreased meaning that .375% of the lamb was decreased

8 0

3 years ago