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Hitman42 [59]
3 years ago
15

Help me plz branniest to whoever right

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
8 0
I’m sorry to use this but, you can’t have someone answer it if they don’t know the options for the answers :/
You might be interested in
25.6 % of what number is 21.12 ?
otez555 [7]

Answer:

The value of number is 82.5

Step-by-step explanation:

Given as :

Let The number  = x

∴ 25.6 % of x = 21.12

Or, 25.6 × x = 21.12 × 100

Or, 25.6 × x = 2112

∴   x = \frac{2112}{25.6}

i.e x = 82.5

Hence 25.6 % of 82.5 is 21.12  So , The value of number is 82.5  Answer

7 0
3 years ago
_ is 8% of 200, please help and show work, thanks
qwelly [4]
% is 1/100 part of whole
so 1% is 200/100=2
and 8% is 2*8=16
or
200/100*8=2*8=16
or
200*0.8=16
Answer: 16
4 0
4 years ago
Read 2 more answers
Three times a number increased by 7 is 100. Find the number.
velikii [3]

Answer:

31

Step-by-step explanation:

100-7= 93

93÷3=31

7 0
3 years ago
Read 2 more answers
Given vectors u = ⟨–5, 1⟩ and v = ⟨8, 6⟩, which statement is true for u and v?
lora16 [44]

Answer:

The vectors form an obtuse angle.

Step-by-step explanation:

On a graph, -5, 1 is in the second quadrant and 8, 6 is in the first. The angle about the orgin is obtuse.

Finding the dot product between them results in a negative, which also means the angle is obtuse.

7 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
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