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3 years ago

10

What is (10/3) squared?

2 answers:

Burka [1]3 years ago

8 0

$(\frac{10}{3})^{2} = \frac{10^{2}}{3^{2}} = \frac{100}{9}$

stiks02 [169]3 years ago

3 0

$\hbox{ We know that } (\frac{a}{b} )^2= \frac{a^2}{b^2}$
$(\frac{10}{3} )^2= \frac{10^2}{3^2}=\boxed{\boxed{ \frac{100}{9} }}$

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Let f(x)=x^2+2x+3 . What is the average rate of change for the quadratic function from x=−2 to x = 5?

miskamm [114]

To find the average rate of change of given function f(x) on a given interval (a,b):

Find f(b)-f(a), b-a, and then divide your result for f(b)-f(a) by your result for b-a:

f(b) - f(a)
------------
b-a

Here your function is f(x) = x^2 - 2x + 3. Substituting b=5 and a=-2,
f(5) = 5^2 -2(5)+3 =? and f(-2) = (-2)^2 - 2(-2) + 3 = ?

Calculate f(5) - [ f(-2) ]
------------------ using your results, above.
5 - [-2]

Your answer to this, if done correctly, is the "average rate of change of the function f(x) = x^2+2x+3 on the interval [-2,5]."

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3 years ago

Which pair shows equivalent expressions?

Dmitry_Shevchenko [17]

Which pair shows equivalent expressions?

A.2(2/5x + 2)=2 2/5x + 1

B.2(2/5x + 2)=4/5x + 4

C.2(2/5x + 4)=4/5x + 2

D.2(2/5x + 4)=2 2/5x + 8

Solution:

$2(\frac{2}{5}x + 2)$

Let us distribute 2 inside the parenthesis.

That is, we use distributive property:

a(b+c)=ab+ac

$2(\frac{2}{5}x + 2) =\frac{2*2}{5}x+2*2$

So, $2(\frac{2}{5}x + 2) =\frac{4}{5}x+4$

Answer:Option (b)

$2(\frac{2}{5}x+4)$

Applying distributive property, a(b+c)=ab+ac

$2(\frac{2}{5}x+4) =2*\frac{2}{5} x+2*4$

$2(\frac{2}{5}x+4) =\frac{2*2}{5} x+2*4$

$2(\frac{2}{5}x+4) =\frac{4}{5} x+8$

So, Option (B) is correct.

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3 years ago

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A boy stands 12 metres from the foot of the building and observes the angle of elevation of the top of the building. The height

nikitadnepr [17]

Answer:

What!!

Step-by-step explanation:

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2 years ago

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A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

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3 years ago

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Crazy boy [7]

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3 years ago

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