Lamy can paint 84 portraits in 6 weeks. at this rate , how many portraits can he paint in 2 weeks

Mathematics

1 answer:

lilavasa [31]1 year ago

7 0

Answer:

28 portraits

Step-by-step explanation:

Let's first figure out how many portraits Lamy can paint in 1 week, which is his <u>unit rate</u>. To calculate this, we just have to divide the number of portraits he paints by the amount of time it takes him to paint them.

In this case, the former quantity is 84 portraits, and the latter quantity is 6 weeks, so his unit rate is $\frac{84 \text \: portraits}{6 \text \: weeks}$ = 14 paintings per week.

Now, we know that in 1 week, Lamy can paint 14 portraits. Therefore, since this is a <u>directly proportional relationship</u>, all we have to do to find how many portraits he can paint is 2 weeks is double the unit rate. This is because in a directly proportional relationship, if you multiply one variable by a number, you have to multiply the other by the same number to maintain equality, and here we are multiplying weeks by 2 so we need to multiply paintings by 2 as well.

Thus, Lamy can paint 14 · 2 = 28 paintings in 2 weeks.

Hope this helps!

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How do you do this question?

Ksivusya [100]

Answer:

V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

$\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}$

The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.

$V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\$

$\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)$

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.

5 0

3 years ago

Given: ABCD is a trapezoid, AC ⊥ CD AB = CD, AC=the square root of 75 , AB = 5 Find: AABCD

Ugo [173]

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