<em>Question:</em>
<em>Triangles PQR and RST are similar right triangles. Which proportion can be used to show that the slope of PR is equal to the slope of RT?</em>
Answer:

Step-by-step explanation:
See attachment for complete question
From the attachment, we have that:





First, we need to calculate the slope (m) of PQR
Here, we consider P and R

Where


becomes
--------- (1)
Next, we calculate the slope (m) of RST
Here, we consider R and T

Where


becomes
---------- (2)
Next, we equate (1) and (2)

<em>From the list of given options (see attachment), option A answers the question</em>
Answer:
depends on what kinda cylinder your starting to find a answer to
Answer:
A = 12.56 m²
General Formulas and Concepts:
<u>Pre-Alg</u>
- Order of Operations: BPEMDAS
<u>Geometry</u>
- Area of a Circle: A = πr²
- d = 2r
Step-by-step explanation:
<u>Step 1: Define</u>
d = 4 m
<u>Step 2: Find </u><em><u>r</u></em>
- Substitute: 4 = 2r
- Isolate <em>r</em>: 2 = r
- Rewrite: r = 2 m
<u>Step 3: Find Area</u>
- Substitute: A = π(2)²
- Substitute: A = (3.14)(2)²
- Evaluate: A = (3.14)(4)
- Multiply: A = 12.56 m²
9514 1404 393
Answer:
A. 3×3
B. [0, 1, 5]
C. (rows, columns) = (# equations, # variables) for matrix A; vector x remains unchanged; vector b has a row for each equation.
Step-by-step explanation:
A. The matrix A has a row for each equation and a column for each variable. The entries in each column of a given row are the coefficients of the corresponding variable in the equation the row represents. If the variable is missing, its coefficient is zero.
This system of equations has 3 equations in 3 variables, so matrix A has dimensions ...
A dimensions = (rows, columns) = (# equations, # variables) = (3, 3)
Matrix A is 3×3.
__
B. The second row of A represents the second equation:

The coefficients of the variables are 0, 1, 5. These are the entries in row 2 of matrix A.
__
C. As stated in part A, the size of matrix A will match the number of equations and variables in the system. If the number of variables remains the same, the number of rows of A (and b) will reflect the number of equations. (The number of columns of A (and rows of x) will reflect the number of variables.)
1st Avenue would be more difficult because it’s rise and run is for every one foot forward it is 3 feet up. Meanwhile avenue 16th would start at (3,1) and the rise and run would be for every 3 feet it would go up 1 foot.