1.Choice 4: 10
2. Choice 1: Add 5.2 to both sides if the equation.
Answer:
The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.
Step-by-step explanation:
Statement is incorrect. Correct form is presented below:
<em>The height </em>
<em> of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation </em>
<em>, where </em>
<em> is time in seconds. After how many seconds will the ball be 84 feet above the ground. </em>
We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

(1)
By Quadratic Formula:

,
The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.
Answer:
X=4
Step-by-step explanation:
First you add like terms, in this case 5x and x. Then you have 6x-1=23. You then add 1 from both sides so you end up with 6x=24. This then results in X=4
Answer:
The tube are similar in shape.
The height of the small tub is 5 cm.
The volume of the small tube is 150 cm3.
The volume of the large tub is 500 cm3.
Work out the height of the large tub.
Give your answer correct to 3 significant figures.
Step-by-step explanation: Hope this help(:
<span>Let r(x,y) = (x, y, 9 - x^2 - y^2)
So, dr/dx x dr/dy = (2x, 2y, 1)
So, integral(S) F * dS
= integral(x in [0,1], y in [0,1]) (xy, y(9 - x^2 - y^2), x(9 - x^2 - y^2)) * (2x, 2y, 1) dy dx
= integral(x in [0,1], y in [0,1]) (2x^2y + 18y^2 - 2x^2y^2 - 2y^4 + 9x - x^3 - xy^2) dy dx
= integral(x in [0,1]) (x^2 + 6 - 2x^2/3 - 2/5 + 9x - x^3 - x/3) dx
= integral(x in [0,1]) (28/5 + x^2/3 + 26x/3 - x^3) dx
= 28/5 + 40/9 - 1/4
= 1763/180 </span>