Factored Form:
x^2 - 4x - 5
Simplifying:
x^2 + - 4x + - 5 = 0

Reorder the terms:
- 5 + - 4x + x^2 = 0

Solving for variable " x":
Subproblem 1:
Set the factor ( - 1 + - 1x) equal to Zero and attempt to solve.

Simplifying:
- 1 + - 1x = 0

Solving:
- 1 + - 1x = 0
Move all terms containing x to the left, all other terms to the right

Add 1 to each side of equation:
- 1 + 1 + - 1x = 0 + 1

Combine Like Terms: 0 + 1 = 1
x = - 1

Divide each side by - 1
x = - 1

Simplifying: x = - 1

Subproblem 2: Set the factor (5 + - 1x) equal to Zero attempt to solve

Simplifying:
5 + - 1x = 0

Move all terms containing x to the left, all other terms to the right
Add - 5 to each side of the equation
5 + - 5 + - 1x = 0 + 5

Combine Like terms: 5 + - 5 = 0
0 + - 1x = 0 + - 5
- 1x = 0 + - 5

Combine Like Terms: 0 + - 5 = - 5
- 1x = - 5

Divide each side by - 1
x = 5

Simplifying:
x = 5

Solution:
x = { - 1, 5}

Answer when factored:
(x + 1)(x - 5)

hope that helps!!!

4 0

3 years ago

Read 2 more answers

Find the product 0 . 67 × 0 .3

quester [9]

0.067 =67/100
0.3 =3/10
0.67 * 0.3 = 67/100 *3/10
(67*3) / (100*10)
=201/1000
=0.201

8 0

3 years ago

Read 2 more answers

Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

Substitute the value of $\lambda$ in the parametric equations:

$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$

7 0

3 years ago

PLZ HELP BRAINLIEST

Montano1993 [528]

Answer:

see explanation below

Step-by-step explanation:

You have three pairs of shoes (white, brown, and black), and you have 6 pairs of pants ( brown, blue, grey, white, sand, and black)

We use the Fundamental Counting Principle to find the number of possible pant-shoe combinations:

3 * 6 = 18

then we can find the probability of wearing by random selection brown shoes with either blue or grey pants:

prob (brown shoes & blue pants) + prob (brown shoes & grey pants) =

= 1/18 + 1/18 = 2/18 = 1/9

3 0

3 years ago