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e-lub [12.9K]
3 years ago
10

What is (10/3) squared?

Mathematics
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]."
5 0
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.

8 0
3 years ago
Read 2 more answers
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:

4 0
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}

6 0
3 years ago
Write the following number in standard decimal form four tenths
Crazy boy [7]
0.4 is four tenths. 0.4 = 4/10=2/5
8 0
3 years ago
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