Ask question

Ask question

All categories

2 years ago

7

The amount of airborne particulate pollution p from a power plant depends on the wind speed s, among other things, with the rela

tionship between p and s approximated byp = 64 − 0.01s2.Find the value(s) of s that will make p = 0.

1 answer:

Bond [772]2 years ago

3 0

Answer:

The value of <em>s</em> that will make <em>p</em> = 0 is 80.

Step-by-step explanation:

The equation used to describe the relationship between amount of airborne particulate pollution from a power plant and wind speed is:

$p=64 -0.01s^{2}$

Compute the value(s) of <em>s</em> for which the value of <em>p</em> = 0 as follows:

$p=64 -0.01s^{2}\\0=64 -0.01s^{2}\\0.01s^{2}=64\\s^{2}=\frac{64}{0.01}\\ s^{2}=6400\\s=\sqrt{6400}\\s=\pm 80$

Since wind cannot travel in negative speed, the value of <em>s</em> is 80.

You might be interested in

You plant a rectangular rose garden along the side of your garage. You enclose 3 sides of the garden with 40 feet of fencing. Th

solmaris [256]

Let's assume

length of rectangle =L

width of rectangle =W

You enclose 3 sides of the garden with 40 feet of fencing

so, we get

$L+2W=40$

now, we can solve for L

$L=40-2W$

we know that

area of rectangle is

$A=L*W$

$100=L*W$

now, we can plug

$100=(40-2W)*W$

now, we can solve for W

$-2W^2+40W-100=0$

we can use quadratic formula

$W=\frac{-40+\sqrt{40^2-4\left(-2\right)\left(-100\right)}}{2\left(-2\right)}$

$W=\frac{-40-\sqrt{40^2-4\left(-2\right)\left(-100\right)}}{2\left(-2\right)}$

we can take anyone value ..because both are giving positive value

first dimensions:

$W=2.929$

now, we can find L

$L=40-2*2.929$

$L=34.142$

so, length is 34.142feet

width is 2.929 feet

Second dimensions:

$W=17.071$

now, we can find L

$L=40-2*17.071$

$L=5.858$

so, length is 5.858feet

width is 17.071 feet

6 0

2 years ago

(05.04 MC) At a school carnival, the diameter of the mat of a trampoline is 16 feet and the diameter of its metal frame is 18 fe

xenn [34]

Answer:

Step-by-step explanation:

You're looking for the circumference here, which is the distance around the outside of a circle. The formula is

C = πd or C = 2πr

Since we are given the diameter, we will use that one:

C = (3.14)(18) so

C = 56.5 feet

3 0

2 years ago

Veronica has been saving dimes and quarters she has 110 coins in all and the total value is 18.50 how many dimes and quarters do

Lady_Fox [76]

This problem can be solved by the chicken rabbits method or you can just do simple algebra.

I.) Chicken and rabbits method

First assume all 110 coins are dimes and none are quarters.

We will have a total value of 11 dollars

Now for each dime we switch out for a quarter, we adds 15 cents to the total value.

18.50-11=7.50 dollars

There are 750/15=50 group of 15 cents in the 7 dollars and 50 cents.

This also meant that we need to switch out 50 dimes for 50 quarters.

So we have 50 quarters.

That first method is very good and very quick once you get the hang of it, now I'm going to show you the algebraic way to solve this.

Let's say there are x dimes and y quarters.

Set up equation

x+y=110

10x+25y=1850

Now solve multiply first equation by 10

10x+10y=1100

subtract

15y=750

y=50

Now we set the numbers of quarters to y so the answer is 50 quarters.

I personally recommend using algebra whenever you can because the practice is very important and you will eventually get really fast at setting up and solving equations. The first method is faster in this case but the second is more generalize, hope it helps.

6 0

2 years ago

Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6

san4es73 [151]

Answer:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

Step-by-step explanation:

Given

$f(x)= \frac{9}{3x+ 2}$

$c = 6$

The geometric series centered at c is of the form:

$\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.$

Where:

$a \to$ first term

$r - c \to$ common ratio

We have to write

$f(x)= \frac{9}{3x+ 2}$

In the following form:

$\frac{a}{1 - r}$

So, we have:

$f(x)= \frac{9}{3x+ 2}$

Rewrite as:

$f(x) = \frac{9}{3x - 18 + 18 +2}$

$f(x) = \frac{9}{3x - 18 + 20}$

Factorize

$f(x) = \frac{1}{\frac{1}{9}(3x + 2)}$

Open bracket

$f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}$

Rewrite as:

$f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}$

Collect like terms

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}$

Take LCM

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}$

$f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}$

So, we have:

$f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}$

By comparison with: $\frac{a}{1 - r}$

$a = 1$

$r = -\frac{1}{3}x + \frac{7}{9}$

$r = -\frac{1}{3}(x - \frac{7}{3})$

At c = 6, we have:

$r = -\frac{1}{3}(x - \frac{7}{3}+6-6)$

Take LCM

$r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)$

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n$

Substitute 1 for a

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n$

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n$

Substitute the expression for r

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n$

Expand

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]$

Further expand:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................$

The power series converges when:

$\frac{1}{3}|x - \frac{7}{3}| < 1$

Multiply both sides by 3

$|x - \frac{7}{3}|$

Expand the absolute inequality

$-3 < x - \frac{7}{3}$

Solve for x

$\frac{7}{3} -3 < x$

Take LCM

$\frac{7-9}{3} < x$

$-\frac{2}{3} < x$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

6 0

2 years ago

The yellow boxes on the bottom are the answers. Please help TYSM

sveticcg [70]

Answer:

$1.\ 8p = 64, p =8$

$2. -\dfrac{w}{4} = 8, w = -32$

$3. -5c = -20, c =4$

$4.\ 9m=45, m = 5$

$5. \dfrac{b}{25} = -3, b =-75$

$6. \dfrac{d}{-2} = 20, d=-40$

Step-by-step explanation:

Now let us see the detailed solution of each part:

Part 1:

$8p -15 = 49\\\Rightarrow 8p =49+15\\\Rightarrow 8p = 64\\\Rightarrow p =8$

Part 2:

$\dfrac{-w}{4}+18=26\\\Rightarrow \dfrac{-w}{4} = 26-18\\\Rightarrow \dfrac{-w}{4} = 8\\\Rightarrow -w = 8 \times 4\\\Rightarrow w =-32$

Part 3:

$-5c-15=-35\\\Rightarrow -5c = -35 +15\\\Rightarrow -5c = -20\\\Rightarrow c = 4$

Part 4:

$9m +10=55\\\Rightarrow 9m =55-10\\\Rightarrow 9m =45\\\Rightarrow m =5$

Part 5:

$\dfrac{b}{25}-83=-86\\\Rightarrow \dfrac{b}{25} = -86 +83\\\Rightarrow \dfrac{b}{25} = -3\\\Rightarrow b = -75$

Part 6:

$\dfrac{d}{-2}+85=105\\\Rightarrow \dfrac{d}{-2} = 105-85\\\Rightarrow \dfrac{d}{-2} = 20\\\Rightarrow d=20 \times -2\\\Rightarrow d =-40$

Please refer to attached image for the answers.

7 0

2 years ago