Answer:

General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
<u>Algebra II</u>
- Distance Formula:

Step-by-step explanation:
<u>Step 1: Define</u>
Point (11, 4) → x₁ = 11, y₁ = 4
Point (5, 8) → x₂ = 5, y₂ = 8
<u>Step 2: Find distance </u><em><u>d</u></em>
Simply plug in the 2 coordinates into the distance formula to find distance <em>d</em>
- Substitute in points [Distance Formula]:

- [√Radical] (Parenthesis) Subtract:

- [√Radical] Evaluate exponents:

- [√Radical] Add:

- [√Radical] Simplify:

Answer: x = 64
Step-by-step explanation:
70 = 1/2((5x+10)-(3x-2)
(remove the inner bracket)
70 = 1/2(5x+10-3x+2)
(transfer 1/2 to LHS)
70*2 = 2x + 12
140 = 2x+12
(transfer 12 to LHS)
140 - 12 = 2x
128 = 2x
(transfer 2 to LHS)
128/2 = x
=>x = 64
hope it helps
Answer:
D. y=-2x-6
Step-by-step explanation:
<u><em>First start with what we know....</em></u>
y = -2x + 3 (Slope Intercept Form)
<u><em>Because of this we can eliminate B. </em></u>
<u><em>Parallel means that the lines wouldn't be touching which means they should have the same slope and the only one with the same slope is D. </em></u>
Answer:
Step-by-step explanation:
hello :
use pytagor-rule :
triangle ABC : (x+15)² =17²+y²....(*)
triangle BDC : y² = x²+8²....(**)
by (**) put the value of y² in (*) :
(x+15)² = 17²+x²+8²
x²+30x+15² = 17²+x²+18²
30x +225 = 289+64
30x = 128
x= 128/30
x= 64/15
put this value in (**) calculate y² ......continu
Answer:
A) Dimensions;
Length = 20 m and width = 10 m
B) A_max = 200 m²
Step-by-step explanation:
Let x and y represent width and length respectively.
He has 40 metres to use and he wants to enclose 3 sides.
Thus;
2x + y = 40 - - - - (eq 1)
Area of a rectangle = length x width
Thus;
A = xy - - - (eq 2)
From equation 1;
Y = 40 - 2x
Plugging this for y in eq 2;
A = x(40 - 2x)
A = 40x - 2x²
The parabola opens downwards and so the x-value of the maximum point is;
x = -b/2a
Thus;
x = -40/2(-2)
x = 10 m
Put 10 for x in eq 1 to get;
2(10) + y = 40
20 + y = 40
y = 40 - 20
y = 20m
Thus, maximum area is;
A_max = 10 × 20
A_max = 200 m²