Multiple Choice

The linear function f i with values f(-1) = 10 and f(5) = -1 is f(x)

zhenek [66]

Answer:

Use the two points to compute the slope, m, then use one of the points in the form y=m(x)+b to find the value of b.

Explanation:

The equation for the slope, m, of a line is:

m=y1−y0x1−x0 [1]

The equation f(2)=−1 tells us that x0=2andy0=−1; substitute this into equation [1]:

m=y1−−1x1−2 [2]

The equation f(5)=4 tells us that x1=5andy1=4; substitute this into equation [2]:

m=4−−15−2 [3]

m=53

Substitute 53 for m into the equation y=m(x)+b

y=53x+b [4]

Substitute 2 for x and -1 for y and the solve for b:

−1=53(2)+b

b=−133

Substitute −133 for b in equation [4]:

y=53x−133 [5]

Check:

−1=53(2)−133
4=53(5)−133

−1=−1
4=4

Hope this helps

3 0

3 years ago

In terms of geometric shapes, a slice of pizza closely resembles what?

Tatiana [17]

Answer:

triangle

Step-by-step explanation:

It looks like a traingle

6 0

3 years ago

The rate of change of the downward velocity of a falling object is the acceleration of gravity (10 meters/sec 2) minus the accel

Alexus [3.1K]

Answer:

See below

Step-by-step explanation:

Write the initial value problem and the solution for the downward velocity for an object that is dropped (not thrown) from a great height.

if v(t) is the speed at time t after being dropped, v'(t) is the acceleration at time t, so the the initial value problem for the downward velocity is

v'(t) = 10 - 0.1v(t)

v(0) = 0 (since the object is dropped)

The equation v'(t)+0.1v(t)=10 is an ordinary first order differential equation with an integrating factor

$\bf e^{\int {0.1dt}}=e^{0.1t}$

so its general solution is

$\bf v(t)=Ce^{-0.1t}+100$

To find C, we use the initial value v(0)=0, so C=-100

and the solution of the initial value problem is

$\bf \boxed{v(t)=-100e^{-0.1t}+100}$

what is the terminal velocity?

The terminal velocity is

$\bf \lim_{t \to\infty}(-100e^{-0.1t}+100)=100\;mt/sec$

How long before the object reaches 90% of terminal velocity?

90% of terminal velocity = 90 m/sec

we look for a t such that

$\bf -100e^{-0.1t}+100=90\rightarrow -100e^{-0.1t}=-10\rightarrow e^{-0.1t}=0.1\\-0.1t=ln(0.1)\rightarrow t=\frac{ln(0.1)}{-0.1}=23.026\;sec$

How far has it fallen by that time?

The distance traveled after t seconds is given by

$\bf \int_{0}^{t}v(t)dt$

So, the distance traveled after 23.026 seconds is

$\bf \int_{0}^{23.026}(-100e^{-0.1t}+100)dt=-100\int_{0}^{23.026}e^{-0.1t}dt+100\int_{0}^{23.026}dt=\\-100(-e^{-0.1*23.026}/0.1+1/0.1)+100*23.026=1,402.6\;mt$

3 0

3 years ago

Someone plsss help with this one

True [87]

Answer:

B) The amount of service fee, in dollars

Step-by-step explanation:

Look for words relating to "one time," as that indicates the constant

4 0

3 years ago

A company that manufactures cell phones has been given a quarterly operating budget of $1,176,912.42. The company's quarterly op

amm1812

Answer:

maximum is 16,853 cell phones

≤ 18.853

Step-by-step explanation:

1,176,912.42. per quarter budget

247,638.00 per quarter fixed cost

We subtract then divide the amount cost of phones and ensure its less or rounded down.

1,176,912.42 -247,638 = 929274.42

929274.42 / 55.14 = 16853

16,852 < m ≤ 16,853

6 0

3 years ago