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3 years ago

10

maggie spent 4.50 on cheese and fruit at the farmers market. she bought 1/8 pound of apples, 1/4 pound of pears and 1.25 pounds

of bananas. if fruit costs 0.80 per pound, how much did maggie spend on cheese. need work please! having a hard time with this one!

1 answer:

serious [3.7K]3 years ago

4 0

The produced answer is 5.5

hope it helps

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Work out 50% of £60<br> needed asap

Finger [1]

<span>30 UK£ is the answer

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3 years ago

Given a sphere with a radius of 7.5 ft.

ad-work [718]

Answer:

volume of the sphere is V=1766.25 $ft^{3}$

Step-by-step explanation:

step:1

<u>The volume of the sphere formula is V= $\frac{4 pi r^{3} }{n}$ </u>

<u>step:2</u>

Given radius of the sphere r=7.5 ft

substitute given radius r=7.5 ft value in volume of the sphere formula

$V=[tex]\frac{4 pi r^{3} }{n}$ \\

V=\frac{4(3.14)(7.5)^{3} }{3}[/tex]

V=1766.25 $ft^{3}$

Final answer is V=1766.25 $ft^{3}$

3 0

3 years ago

The sum of three consecutive odd numbers is 147. what is the smallest of the three numbers?

Andrew [12]

Let x be the smallest odd number.
x+(x+2)+(x+4)=147
3x+6=147
x=47
therefore the smallest odd no. is 47.

3 0

3 years ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

3 years ago

An explanation would be appreciated! Will mark BRAINLIEST for best answer! #3

Bad White [126]

Answer:8:24

Step-by-step explanation:

So since it is 1:3 and 2 dozen is 24 so 24 divided by 3 = 8 so now it’s 8:24

5 0

3 years ago