Is this Linear y=4000-70x if it is how do you know

Answers

Answer 1

Answer:

if we rearrange this we get y=-70x+4000 (only swapped them around) which is the same as the linear equation of y=mx+c as m is -70 and c is 4000 so it is linear

Answer 2

Answer: Yes... It goes in a straight line

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Is 2x + y = 10 linear ?

Answers

Answer:

This is a linear equation in slope-intercept form.

Can someone tell me how to evalaute 6!

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Where is the question?

If you're referring to factorials, then,

6! = 6*5*4*3*2*1 = 720

You start with 6 and count your way down to 1, multiplying along the way.

the market price of an articles is rs 4000. if the price of the aricle including13 % vat is rs 3616 . find the dis.. percent given to it

Answers

Step-by-step explanation:

Answer:seeinthepicture.

Ihopeit helps...

1)simplify the expression (2x)^4

2)the perimeter of a rectangular is twice the sum of its length and its width. the perimeter is 16 meters and its length is 2 meters more then twice its width

3)Find the real numbers x and y that make the equation true.

-3 + yi = x + 6i

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$(2x)^4=16x^4$
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$2(l+w)=16\n l=2w+2\n\n 2(2w+2+w)=16\n 6w+4=16\n 6w=12\n w=2\n\n l=2\cdot2+2\n l=6$
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$-3+yi=x+6i\Rightarrow x=-3 \wedge y=6$

LOOK AT PICTURE THEN ANWSER QUESTION, WHOEVER HAS CORRECT ANSWER I WILL MARK BRAINIEST!

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I am pretty sure it is D. It’s definitely greater or equal to so it’s either B or D

Evaluate $\rm (3^2+1)/(3^2-1)+(5^2+1)/(5^2-1)+(7^2+1)/(7^2-1)+\ldots+(101^2+1)/(101^2-1) =$
With step by step explanation !

Answers

It's easier to deal with the symbolic sum (in sigma notation),

$\displaystyle\sum_(k=1)^(50)((2k+1)^2+1)/((2k+1)^2-1)$

Expanding the terms in the fraction, computing the quotient, and decomposing into partial fractions gives

$((2k+1)^2+1)/((2k+1)^2-1) = (4k^2 + 4k + 2)/(4k^2 + 4k)$

$=\frac12*(2k^2 + 2k + 1)/(k^2 + k)$

$=\frac12\left(2+\frac1{k(k+1)}\right)$

$=\frac12\left(2 + \frac1k - \frac1{k+1}\right)$

and it's the latter two terms that reveal a telescoping pattern.

In case you need more details about the partial fraction decomposition, we are looking for coefficients a and b such that

$\frac1{k(k+1)}=\frac ak+\frac b{k+1}$

$1 = a(k+1) +bk =(a+b)k+a$

which gives a = 1, and a + b = 0 so that b = -1.

Our sum has been rearranged as

$\displaystyle\frac12\sum_(k=1)^(50)\left(2+\frac1k-\frac1{k+1}\right)=\sum_(k=1)^(50)1+\frac12\sum_(k=1)^(50)\left(\frac1k-\frac1{k+1}\right)=50+\frac12\sum_(k=1)^(50)\left(\frac1k-\frac1{k+1}\right)$

The remaining telescoping sum is

1/2 [(1/1 - 1/2) + (1/2- 1/3) + (1/3- 1/4) + … + (1/48- 1/49) + (1/49- 1/50) + (1/50 - 1/51)]

and you can see how there are pairs of numbers that cancel, so that the sum reduces to

1/2 [1/1 - 1/51] = 1/2 [1 - 1/51] = 1/2 × 50/51 = 25/51

So, our original sum ends up being

$\displaystyle\sum_(k=1)^(50)((2k+1)^2+1)/((2k+1)^2-1) = 50 + (25)/(51) = \boxed{(2575)/(51)}$

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