Use the distributive property to solve the equation -2(x+5) =4

Will give brainliest please help ASAP

mihalych1998 [28]

The answer is m^2/n so basically the second option.

3 0

3 years ago

Let f(x)=x2+6x−16 . Enter the x-intercepts of the quadratic function in the boxes. and

gizmo_the_mogwai [7]

Answer:

x = - 8, x = 2

Step-by-step explanation:

Given

f(x) = x² + 6x - 16

To find the x- intercepts equate f(x) to zero, that is

x² + 6x - 16 = 0 ← in standard form

(x + 8)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 8 = 0 ⇒ x = - 8

x - 2 = 0 ⇒ x = 2

8 0

3 years ago

Please, I need help in this ??

nignag [31]

Answer:

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

Step-by-step explanation:

$\int\frac{x^{4}}{x^{4} -1}dx$

Adding and Subtracting 1 to the Numerator

$\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx$

Dividing Numerator seperately by $x^{4} - 1$

$\int 1 + \frac{1}{x^{4}-1 }\, dx$

Here integral of 1 is x +c1 (where c1 is constant of integration

$x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx$ ----------------------------------(1)

We apply method of partial fractions to perform the integral

$\frac{1}{(x-1)(x+1)(x^{2}+1)}$ = $\frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}$ ------------------------------------------(2)

$\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}$

1 = $A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)$ -------------------------(3)

Substitute x= 1 , -1 , i in equation (3)

1 = A(1+1)(1+1)

A = $\frac{1}{4}$

1 = B(-1-1)(1+1)

B = $-\frac{1}{4}$

1 = C(i-1)(i+1)

C = $-\frac{1}{2}$

Substituting A, B, C in equation (2)

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }$

On integration

Here $\int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx$

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\frac{1}{4} ln(x-1)$ - $\frac{1}{4} ln(x+1)$ - $\frac{1}{2} arctanx$ + c2---------------------------------------(4)

Substitute equation (4) back in equation (1) we get

$x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2$

Here c1 + c2 can be added to another and written as c

Therefore,

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

4 0

3 years ago

How can I solve this equation 5x^2-2x-6=-2x^2+9x

wel

Add 2x^2 to both sides
7x^2-2x-6=9x

Subtract 9x from both sides
7x^2-11x-6=0

Factor by using slip and slide
x^2-11x-42
(x-14)(x+3)
(x-14/7)(x+3/7)
(x-2)(7x+3)

Then use the Zero Product Property to find roots
x-2=0
x=2

7x+3=0
7x=-3
x=-3/7

Final answer: x=-3/7, x=2

3 0

3 years ago

1. Consider f(x) = 3x - 5

Rzqust [24]

Answer:

3

Step-by-step explanation:

Here we have the function

$f(x)=3x-5$