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erastovalidia [21]
3 years ago
13

Solve for the value of r. (8r+6)° (9r-7)​

Mathematics
1 answer:
lukranit [14]3 years ago
8 0

Answer:

r = 13

Step-by-step explanation:

these angles are vertical and are congruent

8r + 6 = 9r - 7

6 = r - 7

13 = r

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Please help ASAP I hate maths and am so confused
sveta [45]

Answer:

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Step-by-step explanation:

5 0
3 years ago
3 Write an expression<br> to represent:<br> "Twice a number 51<br> increased by 51"
Darya [45]

Answer:

51x 3+51 is the answer from my side

5 0
3 years ago
Eric has 5 CDs that he is going to give away. He lets his best friend choose 3 of the 5 CDs. How many different groups of 3 CDs
Snowcat [4.5K]
Given a group of n object. We want to make a selection of k objects out of the n object. This can be done in 

C(n, k) many ways, where C(n, k)= \frac{n!}{k!(n-k)!},

where k!=1*2*3*...(k-1)*k


Thus, we can do the selection of 3 cd's out of 5, in C(5,3) many ways, 

where 

C(5, 3)= \frac{5!}{3!2!}= \frac{5*4*3*2*1}{3*2*1*2*1}= \frac{5*4}{2}=10

Answer: 10
5 0
3 years ago
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}

3 0
3 years ago
an experiment consists of flipping a fair coin 4 times. What is the probability of obtaining at least one head?
Dvinal [7]
\mathbb P(X\ge1)=1-\mathbb P(X=0)

The probability of getting 0 heads in 4 tosses (or equivalently, 4 tails) is \dfrac1{2^4}=\dfrac1{16}.

So the desired probability is

1-\dfrac1{16}=\dfrac{15}{16}
7 0
3 years ago
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