Simplify -2a(7a^5-5a+5)

HELP Given: ABCD is a parallelogram and AC bisects BD. Prove: AC bisects

34kurt

2) $ABCD$ is a rhombus (a parallelogram with perpendicular diagonals is a rhombus)

3) $\overline{AC}$ bisects $\angle BCD$ (diagonals of a rhombus bisect the angles from which they are drawn)

8 0

2 years ago

There are 160 deer on Manitow Island. The population of deer is growing by approximately 20% per year. Write an equation using d

soldier1979 [14.2K]

I believe the equation would be
160+0.2y

Because 160 deer and you round the 20% into a decimal point.

3 0

3 years ago

Four friends go hiking. They bring snacks, a compass, and 3 1/3

lisov135 [29]

Answer:

Step-by-step explanation:

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5 0

2 years ago

A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measuremen

Darina [25.2K]

Answer:

$100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy$ ft-lbs.

Step-by-step explanation:

Given:

The shape of the tank is obtained by revolving $y=x^2$ about y axis in the interval $0\leq x\leq 3$ .

Density of the fluid in the tank, $D=100\ lbs/ft^3$

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is, $y(0)=0^2=0$

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in $x=3$ in the parabolic equation . This gives,

$H=3^2=9\ ft$

So, the height of top of tank is, $y(3)=H=9\ ft$

Now, 5 ft above 'H' means $H+5=9+5=14$

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

$\Delta y=(14-y)$ ft

Now, area of cross section of the tank is given as:

$A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section$

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have, $y=x^2$

So, $x=\sqrt y$

Therefore, radius, $r=\sqrt y$

Now, area of cross section is, $A(y)=\pi (\sqrt y)^2$

Work done in pumping the contents to 5 feet above is given as:

$W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy$

Plug in all the values. This gives,

$W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}$

7 0

3 years ago

Use the information from the polling agency to determine the sample size needed to construct a 95% confidence interval with a ma

Ber [7]

Answer:

the sample size n= 771

sample proportion is p = 1/2

Step-by-step explanation:

Step1:-

marginal error formula is $\frac{2S.D}{\sqrt{n} }$

but if 'n' is sample size then $n = \frac{1}{(margin error)^{2} }$

Given margin error is 36% = 0.036

$n= \frac{1}{(0.036)^2}$

on simplification , we get n= 777 polling agencies

Step 2:-

$Margin of error = \frac{2\sqrt{p(1-p)} }{\sqrt{n} }$

substitute values , we get

$\frac{36}{100} = \frac{2\sqrt{p(1-p)} }{\sqrt{771} }$

after simplification, we get

$\sqrt{p(1-p)} = \frac{1}{2}$

squaring on both sides, we get

$p(1-p) = \frac{1}{4}$

multiply 'p' inside

$p-p^{2} =\frac{1}{4}$

cross multiplication '4' and simplify

$4p^{2} -4p +1 =0$ ........(1)

finding 'p' value

The equation (1) is of the form $(a-b)^{2} = a^{2} -2ab+b^{2}$

$(2p)^{2} -2(2p)(1)+1^{2} = (2p-1)^2$

$(2p-1)^2 =0$

$p=\frac{1}{2}$

conclusion:-

The sample size is 771

the sample proportion is p = $\frac{1}{2}$

4 0

3 years ago