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Korolek [52]
3 years ago
7

the third grade at millbridge school has 179 students. if there are ten more 4th graders than 3rd graders, and thirty more 5th g

raders than 3rd graders. how many students are in 3rd, 4th, and 5th grade?
Mathematics
1 answer:
weeeeeb [17]3 years ago
3 0

179 students in 3rd grade, 189 4th graders, and 209 5th graders. 577 students total!

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Step-by-step explanation:

750-300=450

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Exercise 3.3.2: Verify that the system x® 0 1 3 3 1 x® has the two solutions 1 1 e 4t and 1 −1 e −2t a) . b) Write down the gene
nikklg [1K]

Answer:

The general solution of the system is

X(t)=A\left(\begin{array}{c}1 &-1 \end{array}\right)e^{-2t} + B\left(\begin{array}{c}1 &1 \end{array}\right)e^{4t}\\

Step-by-step explanation:

To verify that the system X^{'} =\left[\begin{array}{cc}1&3\\3&1\end{array}\right] X has the solutions given,

First, we determine the Eigen Values β of the matrix of the form X^{'} =A XUsing |A-βI|=0, where I is the Identity Matrix, we have:

|A-\beta I|=0

|\left(\begin{array}{cc}1&3\\3&1\end{array}\right )-\left(\begin{array}{cc}\beta &0\\0&\beta \end{array}\right )|=0

Subtracting matrices

\left|\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right |=0

Taking the determinant

(1-\beta)(1-\beta)-(3X3)=0\\1-\beta-\beta+\beta^{2}-9=0\\\beta^{2}-2\beta-8=0\\\beta^{2}-4\beta+2\beta-8=0\\\beta(\beta-4)+2(\beta-4)=0\\(\beta-4)((\beta+2)=0\\\beta=4  or -2

We determine the eigen vector using\left(\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0

When \beta = -2,

\left(\begin{array}{cc}3 &3\\3&3 \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0 which implies that 3c_{11}+3c_{12}=0 and thus  

If c_{11}=1, c_{12}=-1

When \beta = 4,

\left(\begin{array}{cc}-3 &3\\3&-3 \end{array}\right)\left(\begin{array}{c}c_{21} &c_{22} \end{array}\right)=0 which implies that 3c_{21}-3c_{22}=0 and thus  

If c_{21}=1, c_{22}=1

A general solution of the system is given as

X(t)=Ac_{1}e^{\beta _{1}} + Bc_{2}e^{\beta _{2}t where the c's are the eigen vectors.

Thus the general solution is

X(t)=A\left(\begin{array}{c}1 &-1 \end{array}\right)e^{-2t} + B\left(\begin{array}{c}1 &1 \end{array}\right)e^{4t}\\

7 0
3 years ago
Answer fast please important
steposvetlana [31]

Answer:

c) does not make sense, y-int is (0,-8). a) the graph does show a linear line. b) sure it works, the x-int is 3. d) yes the line is infinite both ways (x,y) x (infinite, -infinite) y (infinite, - infinite)

Step-by-step explanation:

I hope this helps

4 0
3 years ago
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