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LekaFEV [45]
3 years ago
13

Which of these equations have no solution? Check all that apply.

Mathematics
2 answers:
Rus_ich [418]3 years ago
7 0

Answer:

A. 2(x + 2) + 2 = 2(x + 3) + 1

B. 2x + 3(x + 5) = 5(x – 3)

E. 5(x + 4) – x = 4(x + 5) – 1

These answers are correct I took the test

Hope it helps!

Llana [10]3 years ago
3 0
2(x + 2) + 2 = 2(x + 3) + 1
2x + 4 + 2 = 2x + 6 + 1
2x + 6 = 2x + 7 - NO SOLUTION
(subtract 2x from both sides, then 6 = 7 FALSE)

2x + 3(x + 5) = 5(x – 3)
2x + 3x + 15 = 5x - 15
5x + 15 = 5x - 15 - NO SOLUTION

4(x + 3) = x + 12
4x + 12 = x + 12 - ONE SOLUTION

4 – (2x + 5) = (–4x – 2)
4 - 2x - 5 = -4x - 2
-2x - 1 = -4x - 2  - ONE SOLUTION

5(x + 4) – x = 4(x + 5) – 1
5x + 20 - x = 4x + 20 - 1
4x + 20 = 4x + 19 - NO SOLUTION




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To simplify 21/99, divide both sides by 3, to get 7/33.
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4 years ago
Determine the distance between points A (3, 5) and B (2, 1) rounded to the nearest tenth.
sasho [114]

Answer:

4

Step-by-step explanation:

gradient=vertical distance /Horizontal distance

G=V/H

G = -4/-1

G =4

3 0
2 years ago
What is the answer to 5/6 - 3/12=
AnnyKZ [126]
6 x 2 = 12 & 5 x 2 = 10 10 - 3 = 7, therefore 5/6 - 3/12 = 7/12

Remember to always subtract the numerators but never the denominators, same with addition.
7 0
3 years ago
Read 2 more answers
Find the product. (Do not use spaces in your answer.) d j d k
Y_Kistochka [10]
That would be d^2jk.
7 0
3 years ago
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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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