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aleksandr82 [10.1K]
3 years ago
11

What’s the slope of this question

Mathematics
1 answer:
nignag [31]3 years ago
8 0

Answer:

i believe its 1.5

Step-by-step explanation:

its -4.5 crossing over the y-axis then 3 across divide that and get 1.5

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Which of these equations is in POINT-SLOPE form?
goblinko [34]
Y-3=23(x-4) hope this helps
7 0
3 years ago
HELP ASAP PLEASE!!! AND PLEASE SHOW WORK 1. 20% OF ____= 40 2. 30% OF ____=90
lesya692 [45]

Answer: 1. 200, 2. 300

Step-by-step explanation:

40 * 10/2 = 400/2 = 200

or this way:

40 = 20%

40/2 = 20

20 = 10%

20 * 10 = 200

——————————————

90 * 10/3 = 900/3 = 300

or this way:

90 = 30%

90/3 = 30

30 = 10%

30 * 10 = 300

8 0
3 years ago
Read 2 more answers
Based on the family the graph below belongs to, which equation could represent the graph ?
victus00 [196]

Firstly this looks like translated log function graph

  • y=logx

log1=0

As per graph the graph is pulled some inwards y axis but little bit

  • y=log2x

Usually log x leaves y axis at -1 but here it leaves at +2

So

y axis is moved upwards (+3)

Last function

  • y=log(2x)+3

8 0
3 years ago
When Carla looked out at the school parking lot, she noticed that for every 2 minivans , there were 5 other types of vehicles. I
Scilla [17]

Answer:

115 of them are not minivans

Step-by-step explanation:

Because every 7 vehicles has 2 minivans and other vehicles, so 5/7 are not minivans   166x5/2=115

Hope this works! :)

7 0
3 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
4 years ago
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