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densk [106]
3 years ago
15

How are real numbers used to slove problems

Mathematics
1 answer:
Marina86 [1]3 years ago
3 0

Answer:Real numbers are used in measurements of continuously varying quantities such as size and time, in contrast to the natural numbers 1, 2, 3, …, arising from counting. ... The real numbers include the positive and negative integers and fractions (or rational numbers) and also the irrational numbers.

Step-by-step explanation:

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Select two ratio that are equvivalent to 4:18
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Step-by-step explanation:

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If a quadratic equation has a real root, what do you know about the other roots of the equation? Explain.
notsponge [240]

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A Quadratic Equation can have upto 2 roots maximum. So,if one of the roots is a Real number, there are following two possibilities:

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7 0
3 years ago
Given that 2y^3+by-cy+d,where b,c and d are constants,leaves a remainder R when divided by (y+1) , (y-2) and (2y-1). Find the va
Eddi Din [679]

The polynomial remainder theorem says that a polynomial <em>p(x)</em> leaves a remainder of <em>p(k)</em> when it's divided by <em>x</em> - <em>k</em>.

We're given that dividing <em>p(y)</em> = 2<em>y</em>³ + <em>by</em>² - <em>cy</em> + <em>d</em> leaves the same remainder <em>R</em> after dividing it by <em>y</em> + 1, <em>y</em> - 2, and 2<em>y</em> - 1. So we have

<em>p</em>(-1) = 2(-1)³ + <em>b</em>(-1)² - <em>c</em>(-1) + <em>d</em> = <em>R</em>

==>  <em>R</em> = -2 + <em>b</em> + <em>c</em> + <em>d</em>

<em>p</em>(2) = 2(2)³ + <em>b</em>(2)² - <em>c</em>(2) + <em>d</em> = <em>R</em>

==>  <em>R</em> = 16 + 4<em>b</em> - 2<em>c</em> + <em>d</em>

<em>p</em>(1/2) = 2(1/2)³ + <em>b</em>(1/2)² - <em>c</em>(1/2) + <em>d</em> = <em>R</em>

==>  <em>R</em> = 1/4 + <em>b</em>/4 - <em>c</em>/2 + <em>d</em>

<em />

We're also given that <em>y</em> + 2 is a factor, which means dividing <em>p(y)</em> by it leaves no remainder, and so

<em>p</em>(-2) = 2(-2)³ + <em>b</em>(-2)² - <em>c</em>(-2) + <em>d</em> = 0

==>  0 = -16 + 4<em>b</em> + 2<em>c</em> + <em>d</em>

<em />

Solve the system of equations in boldface. You can eliminate <em>d</em> from the first 3 to first solve for <em>b</em> and <em>c</em>, then solve for <em>d</em> :

(-2 + <em>b</em> + <em>c</em> + <em>d</em>) - (16 + 4<em>b</em> - 2<em>c</em> + <em>d</em>) = <em>R</em> - <em>R</em>

-18 - 3<em>b</em> + 3<em>c</em> = 0

<em>b</em> - <em>c</em> = -6

(-2 + <em>b</em> + <em>c</em> + <em>d</em>) - (1/4 + <em>b</em>/4 - <em>c</em>/2 + <em>d</em>) = <em>R</em> - <em>R</em>

-9/4 + 3<em>b</em>/4 + 3<em>c</em>/2 = 0

<em>b</em> + 2<em>c</em> = 3

(<em>b</em> - <em>c</em>) - (<em>b</em> + 2<em>c</em>) = -6 - 3

-3<em>c</em> = -9

<em>c</em> = 3

<em>b</em> - 3 = -6

<em>b</em> = -3

-16 + 4(-3) + 2(3) + <em>d</em> = 0

<em>d</em> = 22

7 0
3 years ago
So right how to do this so help me it hurts
enyata [817]

Answer:

The answer is y=6x+2

Step-by-step explanation:

6 0
3 years ago
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