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Step2247 [10]
3 years ago
7

Help me simplify this

Mathematics
1 answer:
ziro4ka [17]3 years ago
8 0
(x² -5x+4) / ,(x-1). ,=. (,x-1)(x,-4) ,/ (x-1)
= (x-4)
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Plz plz help me i really need help
olya-2409 [2.1K]
Area = πr². So you have the diameter which is 56 millimeters. And the radius (r) is half the diameter, so the radius would be 28 mm. Square 28 would be 784, multiplied by π would equal 2463.00864mm. Rounded to the nearest hundredth would be 2843.01mm².

3 0
3 years ago
Help me a soon as possible I need your help help help please as soon as possible ASAP
evablogger [386]

Answer:

B

Step-by-step explanation:

The answer is b because the lines outside -65 are symbols for absolute value. Absolute value means the distance away from 0, so b would be an appropriate choice.

5 0
3 years ago
Read 2 more answers
6. Lars walks his dog 4 times farther than Michelle
Kobotan [32]

Answer:

0.92

Step-by-step explanation:

7 0
3 years ago
Find the slope of the line passing through the two points.<br><br> (2.1, 3.8), (3.1, 7.6)
bixtya [17]

Answer:

3.8

Step-by-step explanation:

Slope formula- m = y2 - y1            

                                x2 - x1      

x1, y1

2.1, 3.8

                                   7.6 - 3.8

                                    3.1 - 2.1   =   3.8

x2,y2

3.1, 7.6                          

3 0
2 years ago
Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

Therefore,  y=-4x^2  is the equation of this parabola. The graph is also attached.

7 0
3 years ago
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