Answer:
y=3x+4
idk what it has to be equivalent to...I just put it in y-intercept form
Step-by-step explanation:
X=length
y=width
Area (rectangle)=length x width
we suggest this system of equations:
y=x/5
xy=9/20
solve this system of equations by substitution method:
x(x/5)=9/20
x²/5=9/20
Least common multiple=20
4x²=9
x²=9/4
x=⁺₋√(9/4)
we have two solutions:
x₁=-3/2 it does not validate
x₂=3/2 ⇒y=x/5=(3/2)/5=3/10
The dimensions of the lake are:
lengh=3/2 miles
width=3/10 miles
to check:
Area=3/2 miles x 3/10 miles=9/20 miles².
width is 1/5 the length of the lake ⇒ 3/10 miles=(3/2) /5 miles
Answer:
40 in
Step-by-step explanation:
For a width of w, the length is 3w and the area and perimeter are ...
A = LW = (3w)(w) = 3w^2
P = 2(L+W) = 2(3w +w) = 8w
We are given the area, so we can find w to be ...
75 in^2 = 3w^2
25 in^2 = w^2 . . . . . divide by 3
5 in = w . . . . . . . . . square root
Then the perimeter is ...
P = 8w = 8(5 in) = 40 in
Answer:
-2x² - 8x + 15 = 0
Step-by-step explanation:
Given the following algebraic expression;
x² – 7x + 5
-3x² - x + 10
To add the equation together;
x² – 7x + 5 + (-3x² - x + 10) = 0
x² – 7x + 5 - 3x² - x + 10 = 0
Collecting like terms, we have;
(x² - 3x²) - (7x + x) + (5 + 10) = 0
-2x² - 8x + 15 = 0
