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3 years ago

M is directly proportional to r2 When r=2, m=14 Work out the value of m when r=12

Mathematics

1 answer:

Genrish500 [490]3 years ago

8 0

we have M is durectly porpotional to r^2

so M=(k)r^2

and when r=2, m=14

so 14=(k)(2)^2

k=14/4 =7/2

so when r=12

m= (7/2)(12)^2 =(7/2)(144) = 504

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Based on the polynomial reminder theorem, what is the value of the function when x= -5

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Answer:

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Step-by-step explanation:

Given is a polynomial in x

$F (x)= x^4 + 12x^3 + 30x^2 - 12x + 70$

We have to find the remainder when the above polynomial is divided by x+5

Remainder theorem says that f(x) gives remainder R when divided by polynomial x-a means f(a) = R

Applying the above theorem we can say that value of the function when x =-5

= Remainder when f is divided by x+5

= F(-5)

Substitute the value of -5 in place of x

= (-5)^4 + 12(-5)^3 + 30(-5)^2 - 12(-5) + 70

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= 5

Hence answer is 5

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3 years ago

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

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$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

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