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3 years ago

Find the greatest common factor of the following monomials: 39c^2 9c^3'

Mathematics

1 answer:

lana66690 [7]3 years ago

5 0

Answer:

3c^2

Step-by-step explanation:

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My question... please answer asap

konstantin123 [22]

Answer:

slope=1/4

y-intercept=3

y=.25x+3

Step-by-step explanation:

slope=rise over run so

slope=1/4

y-intercept is the value of y where the line crosses the y axis..

y-intercept=3

In y=mx+b form m is slope and b is y-intercept so...

y=.25x+3

6 0

3 years ago

3. for each item, decide whether or not the given expression is defined. for each item that is defined, compute the result. (a)

Sati [7]

The results of given matrices can be obtained using matrix multiplication.

<h3>Find the results of the given matrices:</h3>

Here in the question it is given that,

A = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$ , B = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ , C = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ , D = $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$ ,

E = $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$ , F = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

We have to find AB, BC, CA, CD, $C^{T} A^{T}$ , F², $BD^{T}$ , $A^{T} A$ and FE.

AB = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$

a₁₁ = 1×2 + (-1)×5 + 2×4 = 5, a₁₂ = 1×(-1) + (-1)×1 + 2×6 = 10, a₁₃ = 1×3 + (-1)×2 + 2×(-2) = -3, a₂₁ = 3×2 + 1×5 + 4×4 = 27, a₂₂ = 3×(-1) + 1×1 + 4×6 = 22, a₂₃ = 3×3 + 1×2 + 4×(-2) = 3

AB = $\left[\begin{array}{ccc}5&10&-3\\27&22&3\end{array}\right]$

BC = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$

a₁₁ = 2×1 + (-1)×(-1) + 3×2 = 9, a₂₁ = 5×1 + 1×(-1) + 2×2 = 8, a₃₁ = 4×1 + 6×(-1) + (-2)×2 = -6

BC = $\left[\begin{array}{ccc}9\\8\\-6\end{array}\right]$

CA, CA is not defined since dimension of the matrices are 3×1 and 2×3

$A^{T}E = \left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right]\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$

a₁₁ = 1×(2-i) + 3×(-i) = 2-4i, a₁₂ = 1x(1+i) + 3×(2+4i) = 7+13i, a₂₁ = -1×(2-i) + 1×(-i) = -2, a₂₂ = -1×(1+i) + 1×(2+4i) = 1+3i, a₃₁ = 2×(2-i) + 4×(-i) = 4-6i, a₃₂ = 2×(1+i) + 4×(2+4i) = 10+18i

$A^{T}E = \left[\begin{array}{ccc}2-4i&7+13i\\-2&1+3i\\4-6i&10+18i\end{array}\right]$

CD = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$

a₁₁ = 1×2 = 2, a₁₂ = 1×(-2) = -2, a₁₃ = 1×3 = 3, a₂₁ = -1×2 = -2, a₂₂ = -1×(-2) = 2, a₂₃ = -1×3 = -3,a₃₁= 2×2 = 4, a₃₂ = 2×(-2) = -4, a₃₃ = 2×3 = 6

CD = $\left[\begin{array}{ccc}2&-2&3\\-2&2&-3\\4&-4&6\end{array}\right]$

$C^{T} A^{T} =\left[\begin{array}{ccc}1&-1&2\end{array}\right]\left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right]$

a₁₁ = 1×1 + (-1)×(-1) + 2×2 = 6, a₁₂ = 1×3 + (-1)×1 + 2×4 = 10

$C^{T}A^{T}=\left[\begin{array}{ccc}6&10\end{array}\right]$

F² = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

a₁₁ = i×i + (1-3i)×0 = -1,a₁₂ = i×(1-3i) + (1-3i)×(4+i) = 10-10i, a₂₁= 0×i + (4+i)×0 = 0, a₂₂ = 0×(1-3i) + (4+i)×(4+i) = 15+8i

F² = $\left[\begin{array}{ccc}-1&10-10i\\0&15+8i\end{array}\right]$

$BD^{T}=\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]\left[\begin{array}{ccc}2\\-2\\3\end{array}\right]$

a₁₁ = 2×2 + (-1)×(-2) + 3×3 = 15, a₂₁ = 5×2 + 1×(-2) + 2×3 = 14, a₃₁ = 4×2 + 6×(-2) + (-2)×3 = -10

$BD^{T}= \left[\begin{array}{ccc}15\\14\\-10\end{array}\right]$

$A^{T} A=\left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right] \left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$

a₁₁ = 1×1 + 3×3 = 10, a₁₂ = 1×(-1) + 3×1 = 2, a₁₃ = 1×2 + 3×4 = 14, a₂₁ = -1×1 + 1×3 = 2, a₂₂ = -1×(-1) + 1×1 = 2, a₂₃ = -1×2 + 1×4 = 2, a₃₁ = 2×1 + 4×3 = 14, a₃₂ = 2×(-1) + 4×1 = 2, a₃₃ = 2×2 + 4×4 = 20

$A^{T} A=\left[\begin{array}{ccc}10&2&14\\2&2&2\\14&2&20\end{array}\right]$

FE = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$ $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$

a₁₁ = i×(2-i) + (1-3i)×(-i) = -2+i, a₁₂ = i×(1+i) + (1-3i)×(2+4i) = 13-i, a₂₁ = 0×(2-i) + (4+i)×(-i) = 1-4i, a₂₂ = 0×(1+i) + (4+i)×(2+4i) = 4+18i

FE = $\left[\begin{array}{ccc}-2+i&13-i\\1-4i&4+18i\end{array}\right]$

Hence we can obtain the results of the required matrices using matrix multiplication.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Let A = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$ , B = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ , C = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ , D = $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$ , E = $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$ , F = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

For each item, decide whether or not the given expression is defined. for each item that is defined, compute the result.

AB, BC, CA, CD, $C^{T} A^{T}$ , F², $BD^{T}$ , $A^{T} A$ and FE

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2 years ago

A frog is jumping onto a lily pad. Its height, h (in feet), is recorded at various seconds, t, in the

myrzilka [38]

Answer:

I don't know

Step-by-step explanation:

sorry

7 0

3 years ago

MOVIE TICKETS A company operates three movie

love history [14]

Answer:

438+374+512=1324

Step-by-step explanation:

so that means that they sold 1324 tickets

6 0

3 years ago

Dividing by 100 gives the same answer as multiplying by what number?explain,and give an example

Phantasy [73]

Dividing by 100 is the same as multiplying by 1/100 or 0.01, because when you divide by a fraction, in this case 100/1, you actually multiply by the reciprocal, which would be 1/100. For example 4/100 = 4 x 1/100 = 4/100 = 1/25 = 0.04. You can also just move the decimal two place values to the left to divide by 100. For example, 4/100 = 4. /100 = 0.04 = 1/25

3 0

3 years ago