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2 years ago

The team first experiments with changing the position of the curved pit. In the computer program, the

Mathematics

1 answer:

Tju [1.3M]2 years ago

6 0

Answer:

1). Shifted 2 units right.

2). Shifted 2 units left.

3). Shifted 2 units up.

4). Shifted 2 units down.

Step-by-step explanation:

Parent quadratic function is,

y = x²

1). When curved pit of the parent function is shifted 2 units right,

Translated function will be,

y = (x - 2)²

2). The curved pit is shifted 2 units left,

y = (x + 2)²

3). The curved pit is shifted 2 units up,

y = x² + 2

4). The curved pit is shifted 2 units down,

y = x² - 2

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Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

2 years ago

A gym instructor has a total of 60 hand weight in her studio. Some of the hand weights are 3 pound weights and the rest are 5 po

Elden [556K]

Answer:

35 5-lb weights
25 3-lb weights

Step-by-step explanation:

Let t and f represent the numbers of three- and five-pound weights. The problem statement tells us ...

f + t = 60

f - t = 10

Adding these two equations, we get ...

(f +t) +(f -t) = (60) +(10)

2f = 70 . . . . . eliminate parentheses

f = 35 . . . . . . divide by 2

She has 35 5-lb weights, and 25 3-lb weights.

8 0

3 years ago

The ratio of orange drinks to grape drinks in a vending machine is 10:8. There are 40 grape drinks in the machine. How many oran

umka21 [38]

Answer:

50 orange drinks are in the machine.

Step-by-step explanation:

40/8 = 5

5*10 = 50

6 0

2 years ago

Which of the following numbers would have digit 6 at the unit place.

BabaBlast [244]

Answer:

None

Step-by-step explanation:

All given numbers have digit 2 at the unit place, only one of them = 262 has digit 6 which is at the tenth place

The answer is none

7 0

2 years ago

Which type of transformation is shown in the graph?

AysviL [449]

You need to include the actual picture of the graph :)

5 0

2 years ago