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larisa86 [58]
3 years ago
15

5x (2yz + 4y - 3) Use the distributive property to simplify the expression

Mathematics
1 answer:
AveGali [126]3 years ago
8 0

Answer:

10xyz + 20xy - 15x

Step-by-step explanation:

5 x ( 2yz + 4 y - 3)

using the Distributive property multiply the each term by 5x

5x × 2yz + 5x × 4y - 5x × -3

calculate the product

10xyz + 20xy - 15x

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210 Has 3,5,7 As Prime Factors

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PR has endpoints P (12,6) and R (-8,18) find the length of PR to the nearest tenth
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Use distance formula: \sqrt{(x_2- x_1)^2 + (y_2-y_1)^2

\sqrt{(-8-12)^2 + (18-6)^2} =  \sqrt{544} = 23.3
3 0
3 years ago
Read 2 more answers
Using the definition of inverse (Definition 1, on Page 43) and nothing more, show that if A is an invertible matrix and c is a n
elena-s [515]

Answer:

The matrix cA is invertible and its inverse is \frac{1}{c}\cdot A^{-1}.

Step-by-step explanation:

Since the definition of the inverse matrix states that the inverse of matrix A is a matrix B such that:

A\cdot B=B\cdot A=I

we have to assume the form of such matrix. In our case we have the matrix cA, c\neq 0 and so, the constant c must be somehow eliminated from the equation. The most logical way to do so is to include \frac{1}{c} in the inverse. If we choose matrix B to be B=\frac{1}{c}\cdot A^{-1}, we will have this:

cA\cdot \frac{1}{c}\cdot A^{-1}=c\cdot \frac{1}{c}\cdot A\cdot A^{-1}=1\cdot I=I and

\frac{1}{c}\cdot A^{-1}\cdot cA=\frac{1}{c}\cdot c\cdot A^{-1}\cdot A=1\cdot I=I.

We can form the matrix B like this because we know from the text of the problem that the inverse matrix of A exists and that c is a nonzero number.

<u><em>Here is another way to solve this using the formula of the inverse matrix</em></u>

Since we know that the matrix A is invertible, it follows that its determinant is different from zero. Using the formula for the inverse matrix:

A^{-1}=\frac{1}{\det (A)}\cdot \text{Adj} (A)

we will assume the form of an inverse matrix of cA. We need to obtain the formula for the inverse of cA, so we first need to find \det (cA)\ \text{and}\ \text{Adj} (cA). Since the matrix cA is obtained from matrix A by multiplying every term with c, while calculating determinant we have a constant c that can be extracted from every column (or row) in front. Therefore, we have that

\det (cA)=c^n\cdot \det (A).

On the other hand, \text{Adj} (cA) consists of minors of the matrix cA. Therefore, when we extract the constant in front of such (n-1 \times n-1) determinants, we have c^{n-1} in each column (row). Including all this into the formula we have that:

(cA)^{-1}=\frac{1}{c^n\cdot \det (A)}\cdot c^{n-1} \text{Adj } (A)=\frac{1}{c\cdot \det (A)} \cdot \text{Adj} A=\frac{1}{c}\cdot A^{-1}.

6 0
4 years ago
Anyone mind helping out with my math?
Dmitrij [34]

Answer:

no

Step-by-step explanation:

all the numbers are multiples of each two.

For example,

2×2=4,

4×2=8,

8×2=16

So, the line would be linear.

// have a great day //

3 0
3 years ago
What time is it 4:15-15min
Vesnalui [34]
4:00 Minutes would be your answer because 4:15 - :15 = 4. Hope this helps!
6 0
4 years ago
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