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Scorpion4ik [409]

3 years ago

15

A cylindrical container of height 28 cm and diameter 18 cm is filled with water. The water is then poured into another container

with a rectangular base of length 27 cm and width 11 cm. Calculate the depth (height) of the water in the rectangular container. [Take pi to be 22 over 7]

1 answer:

SVETLANKA909090 [29]3 years ago

6 0

Step-by-step explanation:

then poured into another container with a rectangular base of length 27 cm and width 11 cm. Calculate the depth (height) of the water in the rectangular container

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If the initial figure is 10 and then it is dilated with a scale factor of 2 what's the perimeter of dilated figure?

Gelneren [198K]

Answer:

20

Step-by-step explanation:

Initial figure times scale factor = result

10 times 2 = 20

I hope this helps!

8 0

2 years ago

Round to the nearest hundredth if so. Use 3.14 as pi <br>8. r=<br> d =<br> C=78.8 ft

Kryger [21]

Answer: The radius is 12.54 feet, and the diameter is 25.08 feet.

Explanation: By applying the circumference into the equation r = $\frac{C}{2\pi }$ , we get r = $\frac{78.8}{2\pi }$ which simplifies the radius. And since radius is half of diameter, we multiply by 2 to get the diameter as well to solve the other blank in the problem.

Hope this helps! :D

7 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

3 years ago

Write an exponential function to model the situation. A population of 470 animals decreases at an annual rate of 12%.

Tamiku [17]

$\bf \qquad \textit{Amount for Exponential Decay}
\\\\
A=P(1 - r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &470\\
r=rate\to 12\%\to \frac{12}{100}\to &0.12\\
t=\textit{elapsed time}\\
\end{cases}
\\\\\\
A=470(1-0.12)^t\implies A=470(0.88)^t$

5 0

3 years ago

What is the answer for Y=20x + 40

Law Incorporation [45]

Answer:

y=-40 and x is 2

Step-by-step explanation:

idk

4 0

3 years ago