Answer:
7066.66666667
Step-by-step explanation:
Sorry if i'm wrong
1)
2x^2 - 13x - 24 = 0;
the discriminant is : ( - 13 )^2 - 4 * 2 * ( -24 ) = 169 + 192 = 361 = 19^2 => we have two different rational-number solutions ;
2)
[ -2( x + 2 ) - 3( x - 5 ) ] / [ ( x - 5 )( x + 2 ) ] < 0 <=>
( -5x + 11 ) / [ ( x - 5 )( x + 2 ) ] < 0
We have 2 situations :
a) - 5x + 11 < 0 and ( x - 5 )( x + 2 ) > 0 => x∈ ( 11 / 5 , + oo ) and x∈( -oo, - 2 )U
( 5 , + oo ) => x∈( 5, +oo);
b) - 5x + 11 > 0 and ( x - 5 )( x + 2 ) < 0 => x∈(-oo, 11/5) and x∈( -2, 5 ) =>
x∈( -2, 11/5 );
Finally, x∈ U (-2, 11 / 5 ) U ( 5, +oo).
Answer:
Square base dimension = 15 inches
Maximum volume = 7200 inches^3
Step-by-step explanation:
V = x^2y ..... eq 1
Let the square base be x and the height y
Oversize formular is given by
Y + 4x = 92
Y= 92 - 4x .....eq 2
Put eq 2 into eq 1
V = x^2 ( 92 - 4x^3)
V= 92x^2 - 4x^3
Using derivatives
V= 184x - 12x^2
V'= 0 = 184x - 12x^2
X(184 -12x)
X=0
X = 184/12 = 15.33 approximately 15 inches
Maximum Volume = V= 92(15)^2 -5(15)^3
V= 92(225) - 4(3475)
V= 20700 - 13500
V= 7200 inches^3
Answer: C. 100
Step-by-step explanation:
If the large tank is flowing at a constant rate of -6 per minute then we have to know that the water is moving from the tank. So is asking how many water will be left in the tank after 25 minutes.
You could use the equation
y = -6x + 250 where y is the amount of water left in the tank and x is the number of minutes.
So if x is 25 then plot it in and solve for y.
y = -6(25) +250
y = -150 +250
y= 100
Answer:
A
Step-by-step explanation:
If you get the same amount each day then the graph line would be straight.
