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3 years ago

Marlo had $35 but then spent $5m

Mathematics

2 answers:

Step2247 [10]3 years ago

6 0

Answer:

$30

Step-by-step explanation:

All you do is subtract the 5 dollars that he spent from the 35 dollars he already had and you get $30!

Nookie1986 [14]3 years ago

6 0

Answer:

he had $30 left

Step-by-step explanation:

if Marlo has $35 and then gives someone $5 its just 35-5 which is 30

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Each week, a cook purchases 12 pounds of butter. During the last year, the cook has paid as little as $23.04 and at much as $29.

Tatiana [17]

Each week, a cook purchased 12 LBS. of Butter:

During the Last year: (12 Months):

Cook Paid:

Little: $23.04

Much: $29.40, For Butter he or she purchased in a week:

Question: is: what is the Difference between, the Greatest price per pound, and the least price per pound of butter the cook paid within the last year?

EQUATION:

Least Paid / 12 =====> 23.04 /12

Most Paid / 12 ======> 29.40 / 12

Divide:

23.04 / 12 = 1.92 / LB

29.40 / 12 = 2.45 / LB

Now Subtract:

2.45 - 1.92

Answer ======> 0.53 is the difference, between the greatest price per round, and least price per round of butter the cook would have paid within the last year.

Hope that helps!!! : )

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3 years ago

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Her 6. Karen buys a sweatshirt for $30. The sales

Fynjy0 [20]

Answer:

32.1

Step-by-step explanation:

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3 years ago

1. Each spring, Genevieve helps her mom plant flowers. There are 24 rows of

avanturin [10]

Answer:

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Step-by-step explanation:

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Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$