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inessss [21]
3 years ago
7

After the first term in a sequence, the ratio of each term to the preceding term is r. If the first term is a, what is the third

term?
Mathematics
1 answer:
irinina [24]3 years ago
7 0

Answer:

The 3rd term is

ar^2

Step-by-step explanation:

From the given information, the first term of the sequence is

a_1=a

The common ratio of the sequence is

r

The explicit formula for a geometric sequence is :

a_n=a_1(r^{n-1})

To find the third term, we put n=3 to get:

a_n=a(r^{3-1})

This simplifies to:

a_3=ar^2

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 Chocolate bars come in packs of 8 and graham crackers come in packs of 12.  What is the smallest number of chocolate bars and g
Solnce55 [7]

Answer:

3 chocolate bars, and 2 graham crackers

Someone forgot the marshmallows...... :P

Step-by-step explanation:

Chocolate bars = 8 pack

Graham Crackers = 12pack

To have no crackers or chocolate left over, we need to find LCM

Factors of 8:

8, 16, 24, 32, 40, 48, 56, 72....

Factors of 12:

12, 24, 36, 48, 60, 72

The smallest LCM is 24

Chocolate bars:

24/8 = 3

Graham Crackers:

24/12 = 2

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

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3 years ago
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Evalute the integral (sin^3 (x)/cosx)dx
8090 [49]
\int { \frac{sin ^{3} x}{cos x} } \, dx= \\ = \int { \frac{sin ^{2} x*sin x}{cos x} x} \, dx= \\  =  \int { \frac{(1-sin ^{2} x)*sinx}{cos x} } \, dx =
t = cos x,
dt = - sin x dx;
= \int { \frac{-(1- t^{2}) }{t} } \, dt= \\ = \int {  \frac{t ^{2}-1 }{t} } \, dt = \\  =\int {t} \, dt - \int { \frac{1}{t} } \, dt = t^{2}- ln t = \\ = \frac{cos ^{2} x}{2}-ln ( cos x ) + C

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3 years ago
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3 years ago
Please help with the questions in the image
algol13

First integral:

Use the rational exponent to represent roots. You have

\displaystyle \int\sqrt[8]{x^9}\;dx = \int x^{\frac{9}{8}}\;dx

And from here you can use the rule

\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}+C

to derive

\displaystyle \int\sqrt[8]{x^9}\;dx = \dfrac{x^{\frac{17}{8}}}{\frac{17}{8}}=\dfrac{8x^{\frac{17}{8}}}{17}

Second integral:

Simply split the fraction:

\dfrac{3+\sqrt{x}+x}{x}=\dfrac{3}{x}+\dfrac{\sqrt{x}}{x}+\dfrac{x}{x}=\dfrac{3}{x}+\dfrac{1}{\sqrt{x}}+1

So, the integral of the sum becomes the sum of three immediate integrals:

\displaystyle \int \dfrac{3}{x}\;dx = 3\log(|x|)+C

\displaystyle \int \dfrac{1}{\sqrt{x}}\;dx = \int x^{-\frac{1}{2}}\;dx = 2\sqrt{x}+C

\displaystyle \int 1\;dx = x+C

So, the answer is the sum of the three pieces:

3\log(|x|) + 2\sqrt{x} + x+C

Third integral:

Again, you can split the integral of the sum in the sum of the integrals. The antiderivative of the cosine is the sine, because \sin'(x)=\cos(x). So, you have

\displaystyle \int \left( \cos(x)+\dfrac{1}{7}x\right)\;dx = \int \cos(x)\;dx + \dfrac{1}{7}\int x\;dx = \sin(x)+\frac{1}{14}x^2+C

7 0
3 years ago
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