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3 years ago

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At higher elevations water boils at lower temperatures. The temperature at which water boils is called the boiling point of wate

r. At sea level the boiling point of water is 100oC, while at 2 kilometers of elevation (about 6550 feet) the boiling point of water is 94oC. The function that gives the boiling point of water in oC, B
B
, at an elevation of x
x
kilometers is B(x)=100−3.5x
B(x)=100-3.5x
.

Determine the elevations at which the boiling point of water is between 84 oC and 88 oC. Give your answer accurate to 1 decimal place.

The boiling point of water is between 84 oC and 88 oC when the elevation is between

1 answer:

Irina-Kira [14]3 years ago

4 0

Answer:

The boiling point of water is between 84 °C and 88 °C when the elevation is between 3.4 and 4.6 km.

Step-by-step explanation:

You want x such that ...

84 ≤ B(x) ≤ 88

84 ≤ 100 -3.5x ≤ 88 . . . . substitute for B(x)

-16 ≤ -3.5x ≤ -12 . . . . . . . .subtract 100

4.6 ≥ x ≥ 3.4 . . . . . . . . . . .divide by -3.5; this reverses the inequality

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For f(x)=√(2x+1) , find the following:

Wittaler [7]

Part a.

The domain is the set of x values such that $x \ge -\frac{1}{2}$ , basically x can be equal to -1/2 or it can be larger than -1/2. To get this answer, you solve $2x+1 \ge 0$ for x (subtract 1 from both sides; then divide both sides by 2). I set 2x+1 larger or equal to 0 because we want to avoid the stuff under the square root to be negative.

If you want the domain in interval notation, then it would be $\Big[ -\frac{1}{2} , \infty \Big)$ which means the interval starts at -1/2 (including -1/2) and then it stops at infinity. So technically it never stops and goes on forever to the right.

-----------------------

Part b.

I'm going to use "sqrt" as shorthand for "square root"

f(x) = sqrt(2x+1)

f(10) = sqrt(2*10+1) ... every x replaced by 10

f(10) = sqrt(20+1)

f(10) = sqrt(21)

f(10) = 4.58257569 which is approximate

-----------------------

Part c.

f(x) = sqrt(2x+1)

f(x) = sqrt(2(x)+1)

f(x+2a) = sqrt(2(x+2a)+1) ... every x replaced by (x+2a)

f(x+2a) = sqrt(2x+4a+1) .... distribute

we can't simplify any further

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3 years ago

What is the answer to this question

almond37 [142]

Answer:

C. 24 1/4

Step-by-step explanation:

You turn both the lowest and the highest values into improper fractions and subtract the numerators then turn it back into a mixed fraction.

7 0

2 years ago

The monthly charge for a waste collection service is 1830 dollars for 100 kg of waste and 2460 dollars for 135 kg of waste. (a)

Ulleksa [173]

Answer:

$C=18w+30$

Step-by-step explanation:

We are given that The monthly charge for a waste collection service is 1830 dollars for 100 kg of waste

So, $(x_1,y_1)=(100,1830)$

We are also given that The monthly charge for a waste collection service is 2460 dollars for 135 kg of waste.

So, $(x_2,y_2)=(135,2460)$

We are supposed to find a linear model for the cost, C, of waste collection as a function of the number of kilograms, w.

So, we will use two point slope form :

Formula : $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Substitute the values

$y-1830=\frac{2460-1830}{135-100}(x-100)$

$y-1830=18(x-100)$

$y-1830=18x-1800$

$y=18x-1800+1830$

$y=18x+30$

y denotes the cost

x denotes the weight

So, Replace y with C and x with w

$C=18w+30$

So, a linear model for the cost, C, of waste collection as a function of the number of kilograms, w is $C=18w+30$

7 0

2 years ago

What is the apr for a $585 tablet with 12 payments of $50.81

JulsSmile [24]

Answer:

the answer is 4%

Step-by-step explanation:

585/12=48.75

50.81-48.74=2.06

now 2.06 that's the money from interest rate

so now 2.06/50.81=0.04

so the answer is 4%

3 0

3 years ago

Use the Chain Rule to find the indicated partial derivatives. z = x^4 + xy^3, x = uv^4 + w^3, y = u + ve^w Find : ∂z/∂u , ∂z/∂v

k0ka [10]

I'll use subscript notation for brevity, i.e. $\frac{\partial f}{\partial x}=f_x$ .

By the chain rule,

$z_u=z_xx_u+z_yy_u$

$z_v=z_xx_v+z_yy_v$

$z_w=z_xx_w+z_yy_w$

We have

$z=x^4+xy^3\implies\begin{cases}z_x=4x^3+y^3\\z_y=3xy^2\end{cases}$

and

$\begin{cases}x=uv^4+w^3\\y=u+ve^w\end{cases}\implies\begin{cases}x_u=v^4\\x_v=4uv^3\\x_w=3w^2\\y_u=1\\y_v=e^w\\y_w=ve^w\end{cases}$

When $u=1,v=1,w=0$ , we have

$\begin{cases}x(1,1,0)=1\\y(1,1,0)=2\end{cases}\implies\begin{cases}z_x(1,2)=12\\z_y(1,2)=12\end{cases}$

and the partial derivatives take on values of

$\begin{cases}x_u(1,1,0)=1\\x_v(1,1,0)=4\\x_w(1,1,0)=0\\y_u(1,1,0)=1\\y_v(1,1,0)=1\\y_w(1,1,0)=1\end{cases}$

So we end up with

$\boxed{\begin{cases}z_u(1,1,0)=24\\z_v(1,1,0)=60\\z_w(1,1,0)=12\end{cases}}$

3 0

3 years ago