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3 years ago

Use the binomial theorem to show that 7^n + 2 is divisible by 3, where n is a positive integer.

Mathematics

1 answer:

masha68 [24]3 years ago

7 0

By the binomial theorem,

$7^n=(6+1)^n=6^n+n6^{n-1}+\dbinom n26^{n-2}+\cdots+\dbinom n{n-2}6^2+n6^1+1$

Each term containing a factor 6 is divisible by 3, since 3 divides 6. The remaining 1 and 2 sum to 3, which is also divisible by 3.

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Bill is bisecting an angle with technology. Which of the following describes his next step?

mojhsa [17]

Answer:

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Step-by-step explanation:

the first step is correct

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Read 2 more answers

Find all critical numbers of the function <img src="https://tex.z-dn.net/?f=g%28x%29%3Dx%5E%7B%5Cfrac%7B3%7D%7B4%7D%20%7D-2x%5E%

postnew [5]

Differentiate g using the power rule:

$g'(x) = \dfrac34 x^{\frac34-1} - 2\cdot\dfrac14 x^{\frac14-1}$

$g'(x) = \dfrac34 x^{-\frac14} - \dfrac12 x^{-\frac34}$

$g'(x) = \dfrac14 x^{-\frac34} \left(3 x^{\frac12} - 2\right)$

$g'(x) = \dfrac{3\sqrt x - 2}{4 x^{\frac34}}$

The critical points of g occur where g' is zero or undefined.

We have

$g'(x) = 0 \implies 3\sqrt x - 2 = 0 \implies \sqrt x = \dfrac23 \implies \boxed{x = \dfrac49}$

and the derivative is undefined for

$\dfrac1{g'(x)} = 0 \implies 4x^{\frac34} = 0 \implies \boxed{x=0}$

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2 years ago

Can someone answer my question plz

kirill115 [55]

Answer:

Draw a straight line through these two points (0, 0) and (1, 3).

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"Proportional relationship" implies line or curve that goes through the origin; there is no "y-intercept" in this case.

Plot a dot at (0, 0). Next, move your pencil point 1 unit to the right (you'll end up at (1, 0), and then move it from there up to (1, 3). Draw a straight line through these two points (0, 0) and (1, 3).

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3 years ago