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77julia77 [94]
3 years ago
5

How do I solve this inequality?

Mathematics
1 answer:
stiv31 [10]3 years ago
4 0

you work out the equation from 5 to 3 then compare it to 9

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Y=x3 is that a function?
kotykmax [81]

Answer:

yes

Step-by-step explanation:

y(x) is even or odd according as y(−x)=±y(x) . Here, #y(-x)=-(-x)^3=-(-x^3)=x^3=-y(x). So, y is an odd function of x.

5 0
3 years ago
1. Name the circle<br> 2. Name a chord<br> 3. Name a Tangent
aivan3 [116]
1)ABCDF
2)Chord = AG
3)Tangent is AE
7 0
3 years ago
Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one sam
lisov135 [29]
This is a system
x is the number of advanced tickets sold
y is the number of same day tickets sold
once you label your variables then write the equations
x+ y = 45 total tickets sold
20x + 25y = 1050 money taken in
multiply the top equation by -20 and the x will drop out
-20x - 20y = -900
20x + 25y = 1050. Add the equations result is
5y = 150, divide by 5 ,
y or same day tickets cost $30
substitute 30 into first equation
x + 30 = 45 therefore x = 15
4 0
3 years ago
Solve the simultaneous equations<br>8x – 3y = 30<br>3x + y = 7​
hodyreva [135]
Solve for y in 3x+7=y

y=7−3x

Substitute y=7−3x

y=7−3x into 8x−3y=30

8x−3y=30.

17x−21=30

Solve for x

x in 17x−21=30

17x−21=30.

X=3

Substitute x=3 into y=7−3x

y=-2

Therefore,

x=3

y=−2

​

6 0
3 years ago
In general, the eigenvalues of an upper triangular matrix are given by the entries on the diagonal. The same is true for a lower
SCORPION-xisa [38]

Answer:

Verified!

Step-by-step explanation:

Upper or lower triangular matrix does not make any difference in finding eigenvalues because equalizing determinant to zero will lead to the same result.

Let's apply it for 2x2 matix:

A = \left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0\\0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0\\b&c\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0\\-b&\lambda-c\end{array}\right]=(\lambda-a)(\lambda-c)=0

So, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have (\lambda-a)(\lambda-c).

So, eigenvalues are \lambda_1=a,\:\lambda_2=c

Let's apply it for 3x3 matrix:

A = \left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\\lambda I - A = \left[\begin{array}{ccc}\lambda&0&0\\0&\lambda&0\\0&0&\lambda\end{array}\right]-\left[\begin{array}{ccc}a&0&0\\b&c&0\\d&e&f\end{array}\right]\\\\det(\lambda I - A) = det\left[\begin{array}{ccc}\lambda-a&0&0\\-b&\lambda-c&0\\-d&-e&\lambda-f\end{array}\right]=(\lambda-a)(\lambda-c)(\lambda-f)=0

So as above, eigenvalues are entries on the diagonal because zeros in upper side or lower side vanishes the remaining part and only we have (\lambda-a)(\lambda-c)(\lambda-f).

So eigenvalues are \lambda_1=a,\:\lambda_2=c,\:\lambda_3=f

4 0
4 years ago
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