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3 years ago

Simultaneous equations no idea somebody help !

Mathematics

1 answer:

vodka [1.7K]3 years ago

5 0

5r + 6s = 53

25r + 3s = 49

Meaning of simultaneous is to solve by elimination

Multiply the equation 2 by -2

-2( 25r + 3s = 49)

-50r – 6s = -98

Then add the two equation to eliminate s, so the resulting equation is:

-45r =-45

r = 1

substitute r = 1 to equation 1

5(1) + 6s = 53

6s = 53 – 5

6s = 48

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Don put $3,000 in a savings account with an interest rate of 5% for three years. If the interest is compounded annually, how muc

nikklg [1K]

Compounded anually means
if you put x in at y percent then
you calculate like this
if you put in x at y percent then the money you earn is
x +(x times y)=z=first year
second year=z+(z times y)=s
third year=s+(s times y)=t

percent means parts out of 100 so 5%=5/100=0.05
'of' can be translated as multiply
so

3000 +(3000 times 0.05)=3150=fisrt year
3150+(3150 times 0.05)=3307.5=second year
3307.5 +(3307.5 times 0.05)=3472.88=tird year

he will have 3472.88 at end of 3 years

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3 years ago

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Nimfa-mama [501]

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4 years ago

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Consider the matrix A. A = 1 0 1 1 0 0 0 0 0 Find the characteristic polynomial for the matrix A. (Write your answer in terms of

dusya [7]

Answer with Step-by-step explanation:

We are given that a matrix

$A=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

a.We have to find characteristic polynomial in terms of A

We know that characteristic equation of given matrix $\mid{A-\lambda I}\mid=0$

Where I is identity matrix of the order of given matrix

I= $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

Substitute the values then, we get

$\begin{vmatrix}1-\lambda&0&1\\1&-\lambda&0\\0&0&-\lambda\end{vmatrix}=0$

$(1-\lambda)(\lamda^2)-0+0=0$

$\lambda^2-\lambda^3=0$

$\lambda^3-\lambda^2=0$

Hence, characteristic polynomial = $\lambda^3-\lambda^2=0$

b.We have to find the eigen value for given matrix

$\lambda^2(1-\lambda)=0$

Then , we get $\lambda=0,0,1-\lambda=0$

$\lambda=1$

Hence, real eigen values of for the matrix are 0,0 and 1.

c.Eigen space corresponding to eigen value 1 is the null space of matrix $A-I$

$E_1=N(A-I)$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&-1\end{array}\right]$

Apply $R_1\rightarrow R_1+R_3$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]$

Now,(A-I)x=0[/tex]

Substitute the values then we get

$\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

Then , we get $x_3=0$

And $x_1-x_2=0$

$x_1=x_2$

Null space N(A-I) consist of vectors

x= $\left[\begin{array}{ccc}x_1\\x_1\\0\end{array}\right]$

For any scalar $x_1$

$x=x_1\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

$E_1=N(A-I)=Span(\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Hence, the basis of eigen vector corresponding to eigen value 1 is given by

$\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Eigen space corresponding to 0 eigen value

$N(A-0I)=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

$(A-0I)x=0$

$\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

$\left[\begin{array}{ccc}x_1+x_3\\x_1\\0\end{array}\right]=0$

Then, $x_1+x_3=0$

$x_1=0$

Substitute $x_1=0$

Then, we get $x_3=0$

Therefore, the null space consist of vectors

$x=x_2=x_2\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

Therefore, the basis of eigen space corresponding to eigen value 0 is given by

$\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

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3 years ago

Write the following ratio in their simplest form (a)150minutes:1 1/2 hours (b) 150cm:1060mm

spayn [35]

Answer:

a) 5:3

b) 7.5:5.3

Step-by-step explanation:

150min : 90min (1 and a half hour)

10 : 6

5 : 3

150 × 10 mm : 1060 mm

1500 : 1060

750 : 530

7.5 : 5.3

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2 years ago

Given that r||s and q is a transversal, we know that by the [________]. corresponding angles theorem alternate interior angles t

neonofarm [45]

Answer:

alternate interior angles theorem

Step-by-step explanation:

The alternate interior angles theorem states that when two parallel lines are cut by a transversal, the resulting angles produced are a pair of congruent alternate interior angles.

Given the image attached below, both line k and line l are parallel to each other and also, line t is the transversal, therefore the resulting congruent alternate interior angles produced are:

∠ 4 ≅ ∠6, ∠1 ≅ ∠7

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