All categories

3 years ago

EASY 50 POINTS PLEASE

Mathematics

1 answer:

tatyana61 [14]3 years ago

7 0

Answer:

if this us the same person can u plz let me know if that question was right thanks for the 15 points ♡

You might be interested in

What fraction of 1/2 is 1/3 ? Draw a tape diagram to represent the question

sergeinik [125]

Answer:

1/1.5

Step-by-step explanation:

IM JUST SMART

7 0

3 years ago

What is 27/125 in exponential form?

stealth61 [152]

3/5 I think is the answer

8 0

3 years ago

Read 2 more answers

If 10 is reduced by 4 times what will be the answer??<br> I'm bit confused.....help me out....

babunello [35]

I don't really understand the question

3 0

3 years ago

Given: JK tangent, KH=16, HE=12 Find: JK.

Y_Kistochka [10]

Answer: $JK=8$

Step-by-step explanation:

You can observe in the figure that JK is a tangent and KH is a secant and both intersect at the point K. Then, according to the Intersecting secant-tangent Theorem:

$JK^2=KE*KH$

You know that:

$KH=KE+HE$

Then KE is:

$KE=KH-HE$

$KE=16-12$

$KE=4$

Now you can substitute the value of KE and the value of KH into $JK^2=KE*KH$ and solve for JK. Then the result is:

$JK^2=4*16\\JK^2=64\\JK=\sqrt{64}\\JK=8$

7 0

3 years ago

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

4 years ago