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notsponge [240]
4 years ago
7

(0201 LC)

Mathematics
1 answer:
Zigmanuir [339]4 years ago
3 0

Answer:

its c

3 : 7

3/7

Step-by-step explanation:

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If x = 13 then the value of \sqrt{x-5} is
statuscvo [17]

Answer:

2.23

Step-by-step explanation:

take the squareroot of 13-5

(it's been corrected)

8 0
3 years ago
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freda makes 25,000 per year as a medical assistant. after working 6 months she receives a 6% pay raise. what is her new yearly s
gladu [14]

$26,000 because 25,000 × 0.06 = 1,500.

1,500 + 25,000 + 26,000

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What is 2/25 as a percent?
natka813 [3]
2/25 as a percent is 0.08%

Hope this helps:)
4 0
3 years ago
Sarah has 8 3/4 cups of flour each batch takes 1 2/3 cups of flour which is the best answer for how many batches of sarah can ma
ad-work [718]

Answer:  d

Step-by-step explanation:because if look at the problem 8-1 is 7 and 3/4-2/3 is around 1 so it would be 6

3 0
2 years ago
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
3 years ago
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