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3 years ago

Plzz help me with this question

Mathematics

1 answer:

bixtya [17]3 years ago

6 0

Answer:

x = 5

Step-by-step explanation:

We need to find the value of x in this case.

For this, we need to consider that the triangles are similar. Now, we can find it as follows :

$\dfrac{12}{3}=\dfrac{20}{x}\\\\x=\dfrac{20\times 3}{12}\\\\x=5$

So, the value of x is equal to 5.

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1.In a class of 45 pupils ,21play chess and 23 play cards and 5play both games

Mumz [18]

Step-by-step explanation:

(a)

ii.

21 - 5 = 16 play only chess.

iii.

23 - 5 = 18 play only cards

5 play both.

that is 16+18+5 = 39 that play at least one of these games.

so, 45 - 39 = 6 don't play any of these games.

(b)

we have 16+18 = 34 pupils out of 45 that play exactly one game.

so the probabilty to pick one of them is as usual

desired cases / total cases = 34/45 =

= 0.755555555... ≈ 0.76

8 0

2 years ago

What is the y-intercept of y=3-4xy=3−4xy, equals, 3, minus, 4, x?

balu736 [363]

Answer:

y = 3 - 4x

To find the y intercept let x = 0

y = 3 - 4(0)

y = 3 or ( 0 ,3)

Hope this helps.

8 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.