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3 years ago

Pls help me!!!!!!!!!!!!!

Mathematics

1 answer:

solmaris [256]3 years ago

7 0

The answer is the third option

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Based on the polynomial remainder theorem, what is the value of the function when x=−6 ?

Vaselesa [24]

ANSWER

$f( - 6) = 10$

EXPLANATION

The given function is
$f(x) = {x}^{4} + 8 {x}^{3} + 10 {x}^{2} - 7x + 40$

According to the remainder theorem,

$f( - 6) = R$
where R is the remainder when
$f(x)$

is divided by
$x + 6$

We therefore substitute the value of x and evaluate to obtain,

$f( - 6) = { (- 6)}^{4} + 8 {( - 6)}^{3} + 10 { (- 6)}^{2} - 7( - 6)+ 40$

$f( - 6) = 1296 + 8 ( - 216) + 10 (36) - 7( - 6)+ 40$

$f( - 6) = 1296 - 1728 + 360 + 42+ 40$

This will evaluate to,

$f( - 6) = 10$

Hence the value of the function when
$x = - 6$

is
$10$

According to the remainder theorem,
$10$
is the remainder when

$f(x) = {x}^{4} + 8 {x}^{3} + 10 {x}^{2} - 7x + 40$
is divided by
$x + 6$

3 0

2 years ago

Can someone help me asap!! am timed right now

Aneli [31]

Answer:

Step-by-step explanation:

Find the value of the left hand side (LHS)

LHS = 1/4 // 5/6

LHS = 1/4 * 6/5

LHS = 6/20

LHS = 3/10

Now what will make 3/10 on the right?

B: 1/4 * 6/5 = 6/20 = 3/10

a = 4

b = 6

c = 5

7 0

2 years ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago

P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R = 2.

Georgia [21]

Answer:

Step-by-step explanation:

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2 years ago

WILL MARK BRAINLY Ralph Craig ran the 100m for the US in 1912; when did he next compete in the Olympics?

IrinaK [193]

Answer: The 1948 yachting team.

4 0

3 years ago