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3 years ago

Simplify. Remove all perfect squares from inside the square root. √56z^7

Mathematics

2 answers:

Bogdan [553]3 years ago

3 0

Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

Salsk061 [2.6K]3 years ago

3 0

Answer:

$2$ $z^{3}$ $\sqrt{14z}$

Step-by-step explanation:

You might be interested in

A right circular cone has a radius of 8 in. and a slant height of 12 in. These dimensions are multiplied by 6. By what factor is

galben [10]

Answer:

Factor: 36

Step-by-step explanation:

Ratio of sides

1 : 6

Ratio of areas

1² : 6²

1 : 36

8 0

3 years ago

(1 point) a tank contains 1060 l of pure water. a solution that contains 0.06 kg of sugar per liter enters the tank at the rate

kirill115 [55]

(a) There is 0 kg of sugar in the tank at the beginning since it contains pure water at the start. The sugar only comes from the solution.

(b)

$S' = f(t,S) = \left(0.06 \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) - \left(\dfrac{S}{1060} \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) \ \Rightarrow \\ \\ S' = 0.54 \text{ kg}/\text{min} - \dfrac{9S}{1060}$

So yes, you enter S' = 0.54 - (9S/1060)

(c)

$\displaystyle\frac{dS}{dt} = 0.54 - \frac{9S}{1060} \ \Rightarrow\ \frac{dS}{dt} = \frac{572.4 - 9S}{1060}\ \Rightarrow\ \dfrac{dS}{572.4 - 9S} = \frac{1}{1060} dt\ \Rightarrow \\ \\
\int \dfrac{dS}{572.4 - 9S} = \int \frac{1}{1060} dt\ \Rightarrow\textstyle\ -\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t + C \\ \\
S(0) = 0 \ \Rightarrow\ -\frac{1}{9}\ln|572.4 - 0| = \frac{1}{1060}(0) + C\ \Rightarrow\ C = -\frac{1}{9} \ln 572.4$

$-\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t -\frac{1}{9} \ln 572.4\ \Rightarrow \\ \\
\ln|572.4 - 9S| = \ln 572.4 - \frac{9}{1060}t \ \Rightarrow \\ \\
|572.4 - 9S| = e^{\ln 572.4 - 9t/1060}\ \Rightarrow \\ \\
572.4 - 9S= \pm 572.4 e^{-9t/1060}\ \Rightarrow \\ \\
S = \frac{-1}{9}\left(-572.4 \pm 572.4 e^{-9t/1060}\right)$

But only (+) satisfies $S(0) = 0$

$S= -\frac{1}{9}\left(-572.4 + 572.4 e^{-9t/1060}\right) \\ \\
S= 63.6 - 63.6 e^{-9t/1060}\text{ kg}$

Enter in S = 63.6 - 63.6 * e^(-9t/1060)

3 0

3 years ago

Find the volume of a trapezoid with vertices at (2,0), (2,2), (4,0), and (4,4); about the x-axis.

mr Goodwill [35]

The trapezoid have vertices at (2,0), (2,2), (4,0), and (4,4)
and the rotation about the x-axis.
the points (2,0), (4,0) are located on the x-axis
calculate the function y = m x + c using the other vertices (2,2), (4,4)
by substituting in the function y
m = 1 , and c =0
∴ y = x

The volume can be calculated using
$V =\int\limits^a_b { \pi y^2 } \, dx$

∴ $V = \int\limits^4_2 { \pi y^2 } \, dx = \int\limits^4_2 { \pi x^2 } \, dx
$
∴ $V = \pi \ * \frac{x^3}{3} 
$
∴ $V = \frac{ \pi }{3} * (4^3-2^3)= \frac{56 \pi }{3}$

8 0

3 years ago

If a spherical tank 6 m in diameter can be filled with a liquid for $600, find the cost to fill a tank 12 m in diameter.

Cloud [144]

$1,200
Explaination: 6/600=100x12=1,200

7 0

3 years ago

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago