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Nata [24]
4 years ago
6

Help please !!!!!!!!

Mathematics
1 answer:
Ugo [173]4 years ago
8 0

Answer:

y = -4

x = 3

Step-by-step explanation:

Let's first explain when x = 0

Find where x = 0 on the graph. We are finding what the y-value would be. Trace down the y-axis along the vertical line x = 0. Where does the line intersect the y-axis when x = 0? The answer would be -4

The y-intercept is -4, and the x-value for this is 0, of course. So the correct answer for when x = 0 would be y = -4.

Next, we are told to find x when y = -2. Find where y = -2. Now, staying at y = -2, go horizontally along the x-axis at y = -2 until you reach the line. Boom! You should have connected up with the line at around x = 3. Therefore, when y = -2, x = 3

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the cost of building an airplane with 200 seats is about 100$ million if the cost in million of dollars c is proportional to the
Korvikt [17]

Answer:

i dont remeber the exact answer but with the millions get rid of the 0s to add easier

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
12 is 56% of what number?
Viktor [21]

Answer:

21.42857143 = n

Step-by-step explanation:

Is means equals and of means multiply

12 = 56% *n

Change the percent to a decimal

12 = .56*n

Divide by .56

12/.56 = .56n/.56

21.42857143 = n

6 0
3 years ago
Can someone please help?I tried -4 but it was wrong
klasskru [66]

Answer:

4?

Step-by-step explanation:

8 0
3 years ago
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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.

<h2>Why?</h2>

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
3 years ago
S<br>solve for X<br>2x - Px = 5​
maxonik [38]

Answer:

2x - P = 5 \\ x(2 - P) = 5 \\  \boxed{x =  \frac{5}{(2 - P)} }

5 0
3 years ago
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