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Delvig [45]
3 years ago
9

Can someone help me with these questions please

Mathematics
1 answer:
il63 [147K]3 years ago
4 0

Answer:

Please see the attached pictures for full solution.

For the first question, the inequality sign is \geqslant

because of the word "at least". This means that the profit can be greater or equal to $980.

2nd question part b: I saw your working on another post... you're almost there! After finding the circumference of the semi circle, add the diameter of the semicircle (2 radius) since it's part of the garden too.

For the 3rd question part b, I multiply 12.8 by 15 because we know that from part a that 1cm of the scale drawing represents 15 feet of the actual parking lot. Since the length of the parking lot on the scale drawing is given to be 12.8cm, the actual length is 12.8×15= 192 feet. (I answered this question on another post of yours but I decided to explain it again here)

If you have any questions, feel free to ask! :)

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Amir can run 0.25 kilometers per minute.
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Smh Im stuck plz help me :0
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Answer: don't know sorry

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a jewelry store is having a sale. A ring is now reduced to £840. This is a saving of 40% of the original price. Work out the ori
Helen [10]

£1400

A reduction of 40% means that £840 is 60% of the original price

Divide £840 by 60 to find 1% then multiply by 100 to find original price

original price = £840 × \frac{100}{60} = £1400


7 0
3 years ago
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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

Observe that, because n \geq 2 and is an integer,

n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5

In concusion, the statement is true if and only if n is a non negative integer such that n\leq 77

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  (4n)^4- \sum^{n}_{i=0} (2i)^4 for 77 and 78 you will obtain:

(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064

(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992

7 0
3 years ago
3) There were 35 children and 10 adults at a barbeque.
Mila [183]

Answer:

3. a. 10:35= 2:7

b.35:45=7:9

c.40:50=4:5

7 0
3 years ago
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