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3 years ago

A student was asked to add the fractions below. 33 + 1 For their first step the student wrote the problem as 9/24 + 10/24

Mathematics

1 answer:

Tomtit [17]3 years ago

7 0

Answer:

see below

Step-by-step explanation:

a)The least common multiple of 8 and 12 is 24

3*8 =24

2*12 = 24

3/8 * 3/3 =9/24

5/12 * 2/2 = 10/24

9/24 + 10/24 =19/24

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which of the following best describes the bases of a cylinder? A. Congruent B. Polygons C. Parallel D. Discs (Check All That App

mrs_skeptik [129]

Answer:

A. Congruent and D. Discs

Step-by-step explanation:

You won't see a cylinder that doesn't have congruent bases

Look at the shape of the bases and look at a disc compare their shape

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3 years ago

Б) 3х2 – 10 = 152<br>how to solve with square root

Tju [1.3M]

Answer:

$3\sqrt{6}$ and $-3\sqrt{6}$

Step-by-step explanation:

We will take numbers to one side and variables to one side eliminate the "Square" with "Square Root" and solve for the variable x. The process is shown below:

$3x^2-10=152\\3x^2=152+10\\3x^2=162\\x^2=\frac{162}{3}\\x^2=54\\x=\sqrt{54}$

Now, we need to reduce the radical by using the rule shown below:

$\sqrt{x*y}=\sqrt{x} \sqrt{y}$

Now,

$\sqrt{54}=\sqrt{6*9} =\sqrt{6} \sqrt{9} =3\sqrt{6}$

The correct answer is $3\sqrt{6}$ and $-3\sqrt{6}$

<u>Note:</u> squaring positive and negative makes the same answer

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2 years ago

What is the expanded equivalent of sin(a + b)

laila [671]

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3 years ago

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Stolb23 [73]

The answer is 3! I hope this helps. I used a calculator to get the answer.

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3 years ago

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Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

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2 years ago