1 gal = 400 ft²
2 3/8 gal = ? ft²
to solve, use similar ratios with units gal/ft² and compare
(1)/(400) = (2 3/8)/(x)
*multiply both sides by (x) and divide both sides by (1)/(400), you get:
x = (400)*(2 3/8)/(1) = 400*(2 3/8) = 400*2 + 400*3/8 = 800 + 150 = 950 ft²
<u><em>answer is 950 ft²</em></u>
Answer:
(x,y)→(y,-x)
Step-by-step explanation:
Parallelogram ABCD:
A(2,5)
B(5,4)
C(5,2)
D(2,3)
Parallelogram A'B'C'D':
A'(5,-4)
B'(4,-5)
C'(2,-5)
D'(3,-2)
Rule:
A(2,5)→A'(5,-2)
B(5,4)→B'(4,-5)
C(5,2)→C'(2,-5)
D(2,3)→D'(3,-2)
so the rule is
(x,y)→(y,-x)
This is the concept of sinusoidal, to solve the question we proceed as follows;
Using the formula;
g(t)=offset+A*sin[(2πt)/T+Delay]
From sinusoidal theory, the time from trough to crest is normally half the period of the wave form. Such that T=2.5
The pick magnitude is given by:
Trough-Crest=
2.1-1.5=0.6 m
amplitude=1/2(Trough-Crest)
=1/2*0.6
=0.3
The offset to the center of the circle is 0.3+1.5=1.8
Since the delay is at -π/2 the wave will start at the trough at [time,t=0]
substituting the above in our formula we get:
g(t)=1.8+(0.3)sin[(2*π*t)/2.5]-π/2]
g(t)=1.8+0.3sin[(0.8πt)/T-π/2]
The domain would be x ≥ 0.
This is because the outlet cannot have profit before it was open. Therefore, the growth must be from year 0 to present. If they give a year as starting, you can have an upper limit too, but there is not enough information here to determine that information.
<span>The discriminant of a quadratic equation is the b^2-4ac portion that the square root is taken of. If the discriminant is negative, then the function has 2 imaginary roots, if the discriminant is equal to 0, then the function has only 1 real root, and finally, if the discriminant is greater than 0, the function has 2 real roots. So let's look at the equations and see which have a positive discriminant.
f(x) = x^2 + 6x + 8
6^2 - 4*1*8
36 - 32 = 4
Positive, so f(x) has 2 real roots.
g(x) = x^2 + 4x + 8
4^2 - 4*1*8
16 - 32 = -16
Negative, so g(x) does not have any real roots
h(x) = x^2 – 12x + 32
-12^2 - 4*1*32
144 - 128 = 16
Positive, so h(x) has 2 real roots.
k(x) = x^2 + 4x – 1
4^2 - 4*1*(-1)
16 - (-4) = 20
Positive, so k(x) has 2 real roots.
p(x) = 5x^2 + 5x + 4
5^2 - 4*5*4
25 - 80 = -55
Negative, so p(x) does not have any real roots
t(x) = x^2 – 2x – 15
-2^2 - 4*1*(-15)
4 - (-60) = 64
Positive, so t(x) has 2 real roots.</span>
