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2 years ago

In the right triangle shown in the diagram below,

Mathematics

1 answer:

adelina 88 [10]2 years ago

5 0

QUESTION :-

what is the value of x to the nearest whole number?

ANSWER :-

$\cos( \theta) = \frac{base}{hypotenuse} \\ \cos(30) = \frac{x}{24} \\ \frac{ \sqrt{3} }{ \cancel2} = \frac{x}{ \cancel{24}} = \frac{x}{12} \\ 1.732 = \frac{x}{12} \\ 12 \times 1.732 = x \\ x = 20.7$

the nearest whole number-->

20.7 --> approximately --> 21 units

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In an examination, 80 students secured

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Number of students who secured first class in English only is 20

Step-by-step explanation:

Step 1: Find the number of students who secured first class in English only.

Total number of students who secured first class in English or Maths, T = 80

Number of students who secured first class in Maths, M = 50

Number of students who secured first class in both English and Maths, EM = 10

⇒ Let the number of students who secured first class in English only be x.

Total students with first class, T = M + EM + x

⇒ 80 = 50 + 10 + x

⇒ 80 = 60 + x

∴ x = 80 - 60 = 20

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Step-by-step explanation:

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3 years ago

The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati

makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

$y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\$

And the second one would be

$y = 2-x \,\,\,\,\,, 1 \leq x \leq 2$ .

If you rotate the first region around the "y" axis you get that

$A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51$

And if you rotate the second region around the "y" axis you get that

$A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188$

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first region around the x axis you get that

$A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708$

And if you rotate the second region around the x axis you get that

$A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472$

And the sum would be 1.5708+1.0472 = 2.618

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3 years ago