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1 year ago

The dimensions of a rectangles are 8cm and 12cm. What is the ratio of its length to its perimeter?

Mathematics

1 answer:

kari74 [83]1 year ago

3 0

Answer:

Dimensions = 8 CM and 12 CM

length = 12 CM

width = 8 cm

perimeter = 2 ( length + width)

= 2( 12 + 8)

= 2 ( 20)

= 40 CM

ratio = 12/40

= 6/20

= 3/10

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Use average rates of change to determine if the function in the table is Linear or Non-linear.

nevsk [136]

We have been given a table of values of a function. We are asked to determine whether the given function is linear or nonlinear.

We know that a function is linear when its rate of change (slope) is constant.

Let us find slope for each of the given points using slope formula.

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{-3-0}{4-1}=\frac{-3}{3}=-1$

Similarly, we will find the slopes using other given coordinates.

$m=\frac{-6-(-3)}{7-4}=\frac{-6+3}{3}=\frac{-3}{3}=-1$

$m=\frac{-9-(-6)}{10-7}=\frac{-9+6}{3}=\frac{-3}{3}=-1$

Since the rate of change for each set of points is $-1$ , so the rate of change is constant.

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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

2 years ago

Read 2 more answers

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Allisa [31]

First we know that,

3x+15=5x-65 because of vertical angle thr.

y=y because of reflex property

there are 360 degrees in a circle so..

3x+15+5x-65+2y=360

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so each 2 of the angles are...(40*3+15)=135

Then your new equation is

135+135+2y=360

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2 years ago

A limited-edition poster increases in value each year with an initial value of $18. After 1 year and an increase of 15% per year

Roman55 [17]

Answer:

The required equation is $y = 18(1.15)^x$ .

Step-by-step explanation:

Consider the provided information.

The Initial value of poster = $ 18

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Now verify this by substituting x=1 in above equation.

$y = 18(1.15)^1=20.7$

Which is true.

Hence, the required equation is $y = 18(1.15)^x$ .

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2 years ago

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Y_Kistochka [10]

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