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3 years ago

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

2 answers:

4 0

X=6 and x=-8

To solve this, you would treat x-6=0 as one equation, you add 6 to each side to get rid of -6, hence you would get x=6.

Same goes for x+8=0, subtract 8 from each side to get rid of the +8, then you get x=-8

Rudiy273 years ago

4 0

The third option mam.

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Jake plans to take a summer job at the recreation center as a lifeguard. He can be paid according to two different plans.

Jlenok [28]

Payment on Plan A = $300

x is the number of hours Jake works.

Payment on Plan B = Fixed charges + (charge per hour) × (number of hours worked)

= $150 + $6.25(x)

= 150 + 6.25x

Since, the payment on Plan B should be more than Plan A,

150 + 6.25x > 300

6.25x > 300 - 150

6.25x > 150

$x > \frac{150}{6.25}$

x > 24

The minimum vale for x which satisfies this inequality is 25.

Hence, x = 25.

5 0

3 years ago

Evaluate:<br> Σ10, 8(27-1) =

vfiekz [6]

Answer:

Step-by-step explanation:

7 0

3 years ago

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Ede4ka [16]

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3 0

3 years ago

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

Find the common factor and factor out <br> 9ab^2-3abc

dsp73

Lets go step by step and factor out each as we go

first lets factor out a 3 since 3 is common in 9 and 3

3(3ab^2 - abc)

now lets factor out a since its common in both

3a(3b^2 - bc)

now lets factor out a b since its common in both

3ab(b - c)

we can't factor anything else out so we are done.

4 0

3 years ago