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3 years ago

Four more than five times a number is less than 54. evaluate this problem situation for values of the number.

Mathematics

2 answers:

inna [77]3 years ago

8 0

5x+4<54
subtract 4 from both sides
5x<50
divide both sides by 5
x<10

vovangra [49]3 years ago

8 0

Subtract 4 from both sides
5x<50
divide both sides by 5
x<10

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Please help me if you can thank you. For question A the answer is 5000 brochures. My question is how did they get that answer?

Sophie [7]

Answer:

Step-by-step explanation:

Company A's equation is 900+0.50x5,000

Company B's equation is 1,500+0.38x5,000

Both equal 3,400 and have to be multiplied by 5,000

8 0

3 years ago

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

4 years ago

What is 2<img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D" id="TexFormula1" title="\frac{1}{5}" alt="\frac{1}{5}" align="

vivado [14]

$2\dfrac{1}{5}+1\dfrac{1}{10}=2+\dfrac{1}{5}+1+\dfrac{1}{10}=(2+1)+\left(\dfrac{1}{5}+\dfrac{1}{10}\right)\\\\=3+\left(\dfrac{1\cdot2}{5\cdot2}+\dfrac{1}{10}\right)=3+\left(\dfrac{2}{10}+\dfrac{1}{10}\right)=3+\dfrac{2+1}{10}=\boxed{3\dfrac{3}{10}}$

7 0

3 years ago

Of a total of 100 buses operated within a particular town, 48 usually run on time and 36 of these buses are owned by Jerry. Of t

miv72 [106K]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Below is similar question:

<span>48 out of 100 private buses usually run on time. Out of the buses that are on time, only 36 are owned by Jerry. Out of the remaining 52 buses that usually run late, only 7 are owned by Jerry. What is the probability that a bus is owned by Jerry, given that it runs on time?
A. 0.58
B. 0.67
C. 0.75
D. 0.81
</span>
The answer is C as <span>36/48 = .75</span>

4 0

3 years ago

Read 2 more answers

In sunlight, a vertical yardstick casts a 1 ft shadow at the same time that a nearby tree casts a 15 ft shadow. How tall is the

kati45 [8]

Answer: A) 45 ft

Step-by-step

The yardstick is 3ft tall since 1 yard = 3ft. It is 3 times as tall as its shadow. The tree has a 15ft long shadow. The tree should also be 3 times as tall as its shadow. The tree is 15*3ft tall, so it is 45ft tall. I attached an image of the diagram I made.

4 0

3 years ago