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3 years ago

What does x equal in the equation 14-3(x+1)=22

Mathematics

1 answer:

Anna11 [10]3 years ago

7 0

Answer:

-11/3

Step-by-step explanation:

14-3(x+1)

14-3x-3=22

11-3x=22

-3x=22-11

-3x=11

x=-11/3

x=-3 2/3

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3 years ago

For a project in her Geometry class, Chee uses a mirror on the ground to measure the

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The geometry technique that Chee uses to find the height of the goalpost is the equal ratio of the corresponding sides of similar triangles.

Correct response:

The eight of the goalpost is approximately 16.91 meters.

<h3>Methods used to calculate the height</h3>

The length Chee is using the mirror to measure = The height of her school's football goalpost

The distance of the mirror from the goalpost = 14.35 meters

The distance on the other side of the mirror Chee steps to = 1.4 meters

The distance from her eyes to the ground = 1.65 meters

Required:

How tall is the goalpost.

Solution:

By using similar triangles relationships, we have;

$\dfrac{1.65}{1.4} = \mathbf{\dfrac{Height \ of \ the \ goalpost, h}{14.35} }$

Which gives;

$h = \dfrac{1.65}{1.4} \times 14.35 = 16.9125 \approx \mathbf{ 16.91}$

The height of the goalpost, h ≈ 16.91 meters

Learn more about similar triangles here:

brainly.com/question/10676220

7 0

2 years ago

Find all the zeros of the equation x^4-6x^2-7x-6=0

rusak2 [61]

Answer:

The zeros are

$x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

Step-by-step explanation:

We have been given the equation x^4-6x^2-7x-6=0

Use rational root theorem, we have

$a_0=6,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:6,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:3,\:6}{1}$

$-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2$

$=\left(x+2\right)\frac{x^4-6x^2-7x-6}{x+2}\\$

$=x^3-2x^2-2x-3$

Again factor using the rational root test, we get

$=\left(x+2\right)\left(x-3\right)\left(x^2+x+1\right)$

Using the zero product rule, we have

$x+2=0:\quad x=-2\\x-3=0:\quad x=3\\x^2+x+1=0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\\\=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

Therefore, the zeros are

$x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

4 0

3 years ago

Read 2 more answers