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Bumek [7]
3 years ago
12

QUICK! Giving brainliest to correct answer

Mathematics
2 answers:
Luba_88 [7]3 years ago
5 0
In this situation dominos is the better deal.
garri49 [273]3 years ago
4 0

Answer:

Dominos is the better deal.

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Find the probability of rolling a 2 or a 5 on a standard dice. *
romanna [79]
The probability of a 2 or a 5 is 50%
3 0
3 years ago
Classwork: Multiply, Add, and Subtract Polyno
V125BC [204]
I hope this was helpful

3 0
1 year ago
Solve for x 16x = 20(12)
marysya [2.9K]

Answer: X=15

Step-by-step explanation: Isolate the variable by dividing each side by factors that don't contain the variable.

Hope this helps!! :)

4 0
3 years ago
Find the inverse of the given​ matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3
BabaBlast [244]

Answer:

A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]

Step-by-step explanation:

We want to find the inverse of A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Subtract row 1 multiplied by 4 from row 2

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Add row 1 multiplied by 4 to row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]

  • Subtract row 2 multiplied by 2 from row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]

  • Divide row 3 by −27

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Add row 3 multiplied by 2 to row 1

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Subtract row 3 multiplied by 11 from row 2

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0
3 years ago
Given points A(-1, -2) and B(2, 4) where AP: BP=1:2, find the locus of point P.​
koban [17]

Answer:

x^2 + 4x + y^2 +8y  =  0

Step-by-step explanation:

Given

A = (-1,-2)

B = (2,4)

AP:BP = 1 : 2

Required

The locus of P

AP:BP = 1 : 2

Express as fraction

\frac{AP}{BP} = \frac{1}{2}

Cross multiply

2AP = BP

Calculate AP and BP using the following distance formula:

d = \sqrt{(x - x_1)^2 + (y - y_1)^2}

So, we have:

2 * \sqrt{(x - -1)^2 + (y - -2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}

2 * \sqrt{(x +1)^2 + (y +2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}

Take square of both sides

4 * [(x +1)^2 + (y +2)^2] = (x - 2)^2 + (y - 4)^2

Evaluate all squares

4 * [x^2 + 2x + 1 + y^2 +4y + 4] = x^2 - 4x + 4 + y^2 - 8y + 16

Collect and evaluate like terms

4 * [x^2 + 2x + y^2 +4y + 5] = x^2 - 4x + y^2 - 8y + 20

Open brackets

4x^2 + 8x + 4y^2 +16y + 20 = x^2 - 4x + y^2 - 8y + 20

Collect like terms

4x^2 - x^2 + 8x + 4x + 4y^2 -y^2 +16y + 8y  + 20 - 20 =  0

3x^2 + 12x + 3y^2 +24y  =  0

Divide through by 3

x^2 + 4x + y^2 +8y  =  0

3 0
3 years ago
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