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3 years ago

How do you write a linear equation given the slope and a point that

Mathematics

1 answer:

Serga [27]3 years ago

6 0

Answer:

point slope form

Step-by-step explanation:

hi! we can use point slope form:

y-y1=m(x-x1)

where m is the slope, x1 is the x-coordinate of the given point, and y1 is the y-coordinate of the given point.

example:

slope is 4

ordered pair is (2,3)

y-3=4(x-2)

y-3=4x-8

y=4x-5

hope this helps!

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Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

The tables represent two linear functions in a system,

Alenkasestr [34]

Answer:

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

First table:

(-4, 26), (0, 10) → b = 10

$m=\dfrac{10-26}{0-(-4)}=\dfrac{-16}{4}=-4$

$\boxed{y=-4x+10}$

Second table:

(-4, 14), (0, 2) → b = 2

$m=\dfrac{2-14}{0-(-4)}=\dfrac{-12}{4}=-3$

$\boxed{y=-3x+2}$

We have the system of equations:

$\left\{\begin{array}{ccc}y=-4x+10&(1)\\y=-3x+2&(2)\end{array}\right\\\\\text{Put (1) to (2):}\\\\-4x+10=-3x+2\qquad\text{subtract 10 from both sides}\\-4x=-3x-8\qquad\text{add 3x to both sides}\\-x=-8\qquad\text{change the signs}\\x=8\\\\\text{Put the value of x to (2):}\\\\y=-3(8)+2\\y=-24+2\\y=-22$