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4 years ago

Solve for y -5x-y<3

Mathematics

1 answer:

valentinak56 [21]4 years ago

8 0

-5x - y < 3.....add y to both sides
-5x < y + 3...subtract 3 from both sides
-5x - 3 < y.....or y > -5x - 3

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What fraction is £1 of 13p?

Rasek [7]

I believe it would be 13p im not for sure though

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4 years ago

Ir AB - 4x + 9. BC = 5x + 2, and AC - 56. then find the value for x, AB, BC.

tiny-mole [99]

Answer:

A B - 4 x + 9 B C = 5 x + 2 = A = 9 x/ B + 2/ B − 9 C

A C − 56 = A = 9 x /B + 2/ B − 9 C

x=A=9x/B+2x/B-9C

I hope this helps

6 0

3 years ago

Read 2 more answers

Find each percent change 40 to 2

Nostrana [21]

The question we need to ask is what percent of 2 is 40?

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We need to subract 100% from any percent change:

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3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago

Find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the large

serious [3.7K]

Answer:

17, 19, 21

Step-by-step explanation:

If we denote the smallest number as x, then the middle number will be x + 2 and the largest number will be x + 2 + 2.

The equation is:

x + (x + 2) = 3 * (x + 2 + 2) - 27

2x + 2 = 3x + 12 - 27

2x - 3x = 12 - 27 - 2

-x = -17

x = 17

The numbers are: 17, 19, 21

Let's check to be sure:

17 + 19 = 36

3 * 21 - 27 = 63 - 27 = 36

6 0

3 years ago