MATHEMATICS COLLEGE

Can someone answer this please!!! I only have 5 mins and u will get Brainly too :)

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

(a+b)h / 2

(9+15)6 / 2

24 x 6 / 2

I hope im right!!

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Answers

Answer:

5 and 13

Step-by-step explanation:

I got it right on edge

Answer:

5 and 13, got in right in edge 2020

Step-by-step explanation:

A swimming pool whose volume is 10 comma 000 gal contains water that is 0.03% chlorine. Starting at tequals0, city water containing 0.001% chlorine is pumped into the pool at a rate of 6 gal/min. The pool water flows out at the same rate. What is the percentage of chlorine in the pool after 1 hour? When will the pool water be 0.002% chlorine?

Answers

Answer:

C(60) = 2.7*10⁻⁴

t = 1870.72 s

Step-by-step explanation:

Let x(t) be the amount of chlorine in the pool at time t. Then the concentration of chlorine is

C(t) = 3*10⁻⁴*x(t).

The input rate is 6*(0.001/100) = 6*10⁻⁵.

The output rate is 6*C(t) = 6*(3*10⁻⁴*x(t)) = 18*10⁻⁴*x(t)

The initial condition is x(0) = C(0)*10⁴/3 = (0.03/100)*10⁴/3 = 1.

The problem is to find C(60) in percents and to find t such that 3*10⁻⁴*x(t) = 0.002/100.

Remember, 1 h = 60 minutes. The initial value problem is

dx/dt= 6*10⁻⁵ - 18*10⁻⁴x = - 6* 10⁻⁴*(3x - 10⁻¹) x(0) = 1.

The equation is separable. It can be rewritten as dx/(3x - 10⁻¹) = -6*10⁻⁴dt.

The integration of both sides gives us

Ln |3x - 0.1| / 3 = -6*10⁻⁴*t + C or |3x - 0.1| = e∧(3C)*e∧(-18*10⁻⁴t).

Therefore, 3x - 0.1 = C₁*e∧(-18*10⁻⁴t).

Plug in the initial condition t = 0, x = 1 to obtain C₁ = 2.9.

Thus the solution to the IVP is

x(t) = (1/3)(2.9*e∧(-18*10⁻⁴t)+0.1)

then

C(t) = 3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴t)+0.1)

If t = 60

We have

C(60) = 10⁻⁴*(2.9*e∧(-18*10⁻⁴*60)+0.1) = 2.7*10⁻⁴

Now, we obtain t such that 3*10⁻⁴*x(t) = 2*10⁻⁵

3*10⁻⁴*(1/3)(2.9*e∧(-18*10⁻⁴t)+0.1) = 2*10⁻⁵

t = 1870.72 s

There are 7 bananas, 10 oranges, and 5 apples and 5 apples in the fruit bowl. What is the ratio of bananas to fruit?

Answers

Answer:

5:27

Step-by-step explanation:

A ball is thrown straight out at 80 feet per second from an upstairs window that's 15 feet off the ground. It turns out that its motion can be broken down into its motion in the x-direction (horizontally) and its motion in the y-direction (vertically), and that those two components can be thought of completely independently. In other words, its horizontal motion has no influence on its vertical motion, and vice versa. I've written the two component functions here: x(t) is its position in the x-direction, and y(t) is its position in the y-direction. Which of the following choices is correct? (Ignore air resistance.) NOTE: Use a reference frame that makes all positions (x and y coordinates) of the projectile positive.

Answers

Answer:

Step-by-step explanation:

Given that ball is thrown vertically with intial velocity 80 ft per second at a height of 15 feet off the ground.

Let angle of the ball thrown with horizontal be A

Then vertical velocity = 80 sinA

HOrizontal velocity = 80 cosA

Acceleration here is negative gravity

In fps system gravity g = -32ft/sec^2

s=ut+1/2at^2

Substitute to get

h(t) = 80cos A t-16t^2 +15 (since initial height is 15 ft)

x(t) = 80 sin A t

$x(t) = 80t \n \n y(t) = -16t^2 + 15$

Factor 6q2 − 41q + 30

Answers

(X-6)(6x-5)
So;
X-6=0
X=6
6x-5=0
X=5/6

a tree needs to be stake with four wires, each wire is attached to the trunk 6 feet above the ground, and then anchored to the ground 8 feet from the base of the tree. How many feet of the wire is needed for 5 trees???

Answers

Answer:

Step-by-step explanation:

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