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3 years ago

What are the solutions to the equation (x – 6)(x + 8) = 0?

Mathematics

2 answers:

Vanyuwa [196]3 years ago

8 0

For this case we must find the solutions of the following equation:

$(x-6) (x + 8) = 0$

The equation is factored, so to find its solutions we have:

$x-6 = 0$

Adding 6 to both sides we have:

$x = 6$

On the other hand:

$x + 8 = 0$

With 8 on both sides:

$x = -8$

Thus, the solutions are:

$x = 6\\x = -8$

ANswer:

$x = 6\\x = -8$

VikaD [51]3 years ago

3 0

Answer:

x=6, x=-8

Step-by-step explanation: i used the symbolab calculator online to solve this.

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PLEASE HELP

LekaFEV [45]

Answer: one solution

Step-by-step explanation:

CONCEPT:

- One solution is when the final variable would be able to find a solution.

- Infinite solution is when the equations result in 0=0, meaning any number will fit

- No solution is when the equation results in both sides on equal.

SOLVE:

Given

3/4x-7=3(1/6x+1)

Expand Parenthesis

3/4x-7=1/2+3

Multiply both sides by the LCM of fractions (Least Common Multiple)

4(3/4x-7)=4(1/2x+3)

3x-28=2x+12

Subtract both sides by 2x

3x-28-2x=2x+12-2x

x-28=12

Add both sides by 28

x-28+28=12+28

x=40

Since we are able to get exactly one solution, there is only one solution.

Hope this helps!! :)

Please let me know if you have any quesionts

5 0

3 years ago

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x

RoseWind [281]

Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

P(a < X < b) = $\int_{a}^{b} \frac{2}{x^{3}} dx$ = $2\int_{a}^{b} x^{-3} dx$

= $2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx$ { Because $\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a$ }

= $2[ \frac{x^{-2} }{-2}]^{b}_a$ = $\frac{2}{-2} [ x^{-2} ]^{b}_a$

= $-1 [ b^{-2} - a^{-2} ]$ = $\frac{1}{a^{2} } - \frac{1}{b^{2} }$

a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }

Comparing with general probability we get,

P(1 < X < 5) = $\frac{1}{1^{2} } - \frac{1}{5^{2} }$ = $1 - \frac{1}{25}$ = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/ $8^{2}$ - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = $\frac{1}{6^{2} } - \frac{1}{10^{2} }$ = $\frac{1}{36} - \frac{1}{100 }$ = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

= $(\frac{1}{1^{2} } - \frac{1}{6^{2} })$ + (1/ $10^{2}$ - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

⇒ P(1 < X < x) = 0.75

⇒ $\frac{1}{1^{2} } - \frac{1}{x^{2} }$ = 0.75

⇒ $\frac{1} {x^{2} }$ = 1 - 0.75 = 0.25

⇒ $x^{2}$ = $\frac{1}{0.25}$ ⇒ $x^{2}$ = 4 ⇒ x = $2$

Therefore, value of x such that P(X < x) = 0.75 is 2.

8 0

3 years ago