You would earn $15,000 in 500 mins
First you want to find how many “10 seconds” are in 500mins. To do this convert 500 mins to seconds by multiplying 500 by 60 which will give you 30,000. Next divide 10 by 30,000 which will be 3,000. With that you can multiply by 5 to get how much you would earn in 500 mins. 5*3,000=15,000.
The commutative property<span />
Answer:
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Step-by-step explanation:
The interpretation of the given question is as follows:
Use the given inverse to solve the system of equations

The inverse of ![\left[\begin{array}{ccc}1&-1&1\\0&2&1\\3&-8&0\end{array}\right] is \left[\begin{array}{ccc}-8&8&3\\-3&3&1\\6&-5&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-1%261%5C%5C0%262%261%5C%5C3%26-8%260%5Cend%7Barray%7D%5Cright%5D%20%20%20is%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-8%268%263%5C%5C-3%263%261%5C%5C6%26-5%26-2%5Cend%7Barray%7D%5Cright%5D)
x =
y =
z =
Answer:
x = - 1.5
y = - 0.5
z = - 5
Step-by-step explanation:
Using the correlation of inverse of matrix AX = B to solve the question above;
AX = B
⇒ A⁻¹(AX) = A⁻¹ B
X = A⁻¹ B
So ;
X = A⁻¹ B
![\left[\begin{array}{ccc}-6\\ -6\\- \dfrac{1}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-6%5C%5C%20-6%5C%5C-%20%5Cdfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(-8*-6)+(8*-6)+(3*-\dfrac{1}{2})\\(-3*-6)+(3*-6)+(1*-\dfrac{1}{2})\\(6*-6)+(5*-6)+(-2* - \dfrac{1}{2})\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%28-8%2A-6%29%2B%288%2A-6%29%2B%283%2A-%5Cdfrac%7B1%7D%7B2%7D%29%5C%5C%28-3%2A-6%29%2B%283%2A-6%29%2B%281%2A-%5Cdfrac%7B1%7D%7B2%7D%29%5C%5C%286%2A-6%29%2B%285%2A-6%29%2B%28-2%2A%20-%20%5Cdfrac%7B1%7D%7B2%7D%29%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(48)+(-48)+(\dfrac{-3}{2})\\(18)+(-18)+(\dfrac{-1}{2})\\(-36)+(30)+(1)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%2848%29%2B%28-48%29%2B%28%5Cdfrac%7B-3%7D%7B2%7D%29%5C%5C%2818%29%2B%28-18%29%2B%28%5Cdfrac%7B-1%7D%7B2%7D%29%5C%5C%28-36%29%2B%2830%29%2B%281%29%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(\dfrac{-3}{2})\\(\dfrac{-1}{2})\\(-5)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%28%5Cdfrac%7B-3%7D%7B2%7D%29%5C%5C%28%5Cdfrac%7B-1%7D%7B2%7D%29%5C%5C%28-5%29%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}-1.5\\-0.5\\ -5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1.5%5C%5C-0.5%5C%5C%20-5%5Cend%7Barray%7D%5Cright%5D)
Answer:
Δ PQT ~ Δ QRS .....{S-S-S test for similarity}...Proof is below.
Step-by-step explanation:
Given:
In Δ PQT
PQ = 30 ft
QT = 28 ft
TP = 20 ft
In Δ QRS
QR = 15 ft
RS = 14 ft
SQ = 10 ft
To Prove:
Δ PQT ~ Δ QRS
Proof:
First we consider the ratio of the sides
..............( 1 )
..............( 2 )
..............( 3 )
So By equation ( 1 ), ( 2 ) and ( 3 ) we get

Now in Δ PQT and Δ QRS we have

Which are corresponding sides of a similar triangle in proportion.
∴ Δ PQT ~ Δ QRS .....{S-S-S test for similarity}...Proved