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Deffense [45]
3 years ago
13

For what temperature if any, does F=3C?

Mathematics
1 answer:
Vladimir [108]3 years ago
7 0
3 celsius is equal to 37.4 Ferenheight
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Alice make 2 bunches of balloons turre are the same Numbers of red balloons and the same Numbers of blue ballons in each did sir
german
She put 8 balloons in each color group
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3 years ago
For the function y = 15x – 4, assume the domain is only values of x from 0 to 5. What is the range of the function?​
Harrizon [31]
X = 0 y = 15(0) - 4 = -4
x = 1 y = 15(1) - 4 = 11
x = 2 y = 15(2) - 4 = 26
x = 3 y = 15(3) - 4 = 41
x = 4 y = 15(4) - 4 = 56
x = 5 y = 15(5) - 4 = 71
range = largest - smallest
71 - -4 = 75
so the range is 75
3 0
3 years ago
Sistemas de ecuaciones aplicando sustitucion me ayudan porfa
gregori [183]

Answer:

jdbhbehfbhwbfhbwehfbhjwebfwhjfhsbfhbh loool :)

Step-by-step explanation:

5 0
3 years ago
The following line segments are graphed.
bogdanovich [222]

9514 1404 393

Answer:

  (a)  The figure is a quadrilateral.

Step-by-step explanation:

The four segments are not listed in order around the figure, but they do form a closed loop. The figure is a quadrilateral.

4 0
3 years ago
The number of daily sales of a product was found to be given by S = 600xe−x2 + 600 x days after the start of an advertising camp
Elena-2011 [213]

Answer:

a. 20520

b. 12600

Step-by-step explanation:

Given the function S = 600xe^-x² + 600.

a. To find the average, we have to find the definite integral of the function because average, as it is known, is the sum of data points divided by the size of its dataset, this can be used for discrete data. Integral formula is just the continuous form of average, so we are using integral because we were given an interval of x= 0 to X = 30.

Let's integrate 100xe^x² first. Let –x² = u, this means –2xdx = du and we have dx = –du/2x. Also, when x = 0, u = –(0)² = –0 and when x = 30, u = –(30)² = –900. When we make our substitutions we have:

–600(xe^udu)/2x = –600(e^udu)/2 upon integrating that we have –600(e^u)/2. Applying our interval we have

–600[(e^900)/2 – (e^0)/2] ≈ – [– 3.7 – (1/2)] = –600 (–4.2) = 600 x 4.2 = 2520

Now let's integrate 600, with the interval x = 0 to x = 30 (we are using this interval here because the substitution we made didn't affect this).

We have, upon integrating:

600x.

Substituting our intervals we have:

600(30 – 0) = 600 x 30 = 18000.

Adding that up we have: 2520 + 18000 = 20520.

b. The same method is needed, just difference of interval.

Therefore, after integrating the first component with intervals u = 900 to u = 2500 (from x² = u) we have:

–600[(e^2500)/2 – (e^900)/2] ≈ –600[2.7 – 3.7] = –600(–1) = 600.

Then for the second component:

600x using x = 30 to x = 50

600(50 – 30) = 600 x 20 = 12000.

Adding that up we have:

12000 + 600 = 12600.

4 0
3 years ago
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