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max2010maxim [7]

2 years ago

6

A pet supply company makes a 5-gallon aquarium with a 10 inch by 10 inch base and a height of 14 inches. A museum wants to speci

al-order a 135-gallon aquarium with the same shape. Find the new dimensions (length, width and height) of the new aquarium. Show or explain your work.

1 answer:

Nimfa-mama [501]2 years ago

6 0

New dimensions: 72 in for the length, 18 in for the width and 24 in for the height the volume in liters for these dimensions is-509.....but in gallons it’s 135

Hope this helps

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Find a number C such that the polynomial p(x)= -x+4x^2+Cx^3-8x^4

Ksenya-84 [330]

Answer:

C = 2

Step-by-step explanation:

$p(x)= -x+4x^2+Cx^3-8x^4 \\ \\ \because \: given \: polynmial \: has \: zero \: at \: \\x = \frac{1}{4} \\ \\ \implies \: p\bigg(\frac{1}{4} \bigg) = 0...(1) \\ \\ plug \: x = \frac{1}{4} \: in \: p(x) \: we \: find \\ \\ p \bigg(\frac{1}{4} \bigg) = - \frac{1}{4} + 4 {\bigg(\frac{1}{4} \bigg)}^{2} + c {\bigg(\frac{1}{4} \bigg)}^{3} - 8 {\bigg(\frac{1}{4} \bigg)}^{4} \\ \\ 0 = - \frac{1}{4} + \cancel 4 \times {\frac{1}{\cancel{16} } } + c {\bigg(\frac{1}{64} \bigg)} - \cancel 8 \times \frac{1}{\cancel{256} } \\ \\0 =\cancel{ - \frac{1}{4}} + \cancel {{\frac{1}{4} }} + c {\bigg(\frac{1}{64} \bigg)} - \times \frac{1}{32} \\ \\0 = c {\bigg(\frac{1}{64} \bigg)} - \frac{1}{32} \\ \\c {\bigg(\frac{1}{64} \bigg)} = \frac{1}{32} \\ \\ c = \frac{1}{32} \times 64 \\ \\ c = 2$

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3 years ago

What is not a potential source of bias: Of the 2000 students attending, every 5th student going into the basketball game is aske

ahrayia [7]

I think it is C, not a large enough sample.

8 0

3 years ago

Read 2 more answers

The sum of two terms of gp is 6 and that of first four terms is 15/2.Find the sum of first six terms.

Gnoma [55]

Given:

The sum of two terms of GP is 6 and that of first four terms is $\dfrac{15}{2}.$

To find:

The sum of first six terms.

Solution:

We have,

$S_2=6$

$S_4=\dfrac{15}{2}$

Sum of first n terms of a GP is

$S_n=\dfrac{a(1-r^n)}{1-r}$ ...(i)

Putting n=2, we get

$S_2=\dfrac{a(1-r^2)}{1-r}$

$6=\dfrac{a(1-r)(1+r)}{1-r}$

$6=a(1+r)$ ...(ii)

Putting n=4, we get

$S_4=\dfrac{a(1-r^4)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}$

$\dfrac{15}{2}=6(1+r^2)$ (Using (ii))

Divide both sides by 6.

$\dfrac{15}{12}=(1+r^2)$

$\dfrac{5}{4}-1=r^2$

$\dfrac{5-4}{4}=r^2$

$\dfrac{1}{4}=r^2$

Taking square root on both sides, we get

$\pm \sqrt{\dfrac{1}{4}}=r$

$\pm \dfrac{1}{2}=r$

$\pm 0.5=r$

Case 1: If r is positive, then using (ii) we get

$6=a(1+0.5)$

$6=a(1.5)$

$\dfrac{6}{1.5}=a$

$4=a$

The sum of first 6 terms is

$S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}$

$S_6=\dfrac{4(1-0.015625)}{0.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Case 2: If r is negative, then using (ii) we get

$6=a(1-0.5)$

$6=a(0.5)$

$\dfrac{6}{0.5}=a$

$12=a$

The sum of first 6 terms is

$S_6=\dfrac{12(1-(-0.5)^6)}{(1+0.5)}$

$S_6=\dfrac{12(1-0.015625)}{1.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Therefore, the sum of the first six terms is 7.875.

5 0

2 years ago

If 1 card is drawn at random from a standard 52-card deck, what is the probability that the card drawn is a king?

Ahat [919]

Answer:

1/13 chance or roughly 7.7% chance.

Step-by-step explanation:

There are 4 Kings in a standard 52-card deck. If one King was to be pulled out, you'd get a probability of 4/52 chance of pulling a king. Simplified, 4/52 is 1/13 chance.

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3 years ago

What is .181818... as a fraction?

LekaFEV [45]

181818/1000000. let me know if im wrong! <3

4 0

2 years ago