All categories

3 years ago

Solve the equation by factoring plssss help ASAPPPP 4x^{2}-20x=0 Thanks UwU

Mathematics

2 answers:

erastovalidia [21]3 years ago

8 0

X= 0 and 5 should be the right answer

JulijaS [17]3 years ago

4 0

Answer:

x = 0 and 5

Step-by-step explanation:

4x^2 - 20x = 0

Factor 4x

4x(x - 5) = 0

4x = 0 = > x = 0

x- 5 = 0 = > x = 5

You might be interested in

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

3 years ago

A sequence is defined by the recursive formula f (n 1) = f(n) – 2. if f(1) = 18, what is f(5)?

Marat540 [252]

Would assume:

f(n + 1) = f(n) - 2

f(2) = f(1 + 1) = f(1) - 2 = 18 - 2 = 16

f(3) = f(2 + 1) = f(2) - 2 = 16 - 2 = 14

f(4) = f(3 + 1) = f(3) - 2 = 14 - 2 = 12

f(5) = f(4 + 1) = f(4) - 2 = 12 - 2 = 10

f(5) = 10

4 0

3 years ago

Who know the answer

Shkiper50 [21]

They just added 100 each time so i think the answer is b

3 0

3 years ago

When the outliers are removed, how does the mean change?

vagabundo [1.1K]

Answer:

The mean increases by 3.

Step-by-step explanation:

What's an outlier?

An outlier is a data point that's very different from the other data points. In this example, the outlier is 50, as the other data points are around the number 80.

What's the mean of a data set?