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3 years ago

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A vertical plate is submerged (or partially submerged) in water and has the indicated shape. Explain how to approximate the hydr

ostatic force against one side of the plate by a Riemann sum. Then express the force as an integral and evaluate it.

1 answer:

Ivan3 years ago

4 0

Answer:

f = 9.04 x 10 ∧5

Step-by-step explanation:

check the attachment below for detailed explanations

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◆ Quadratic Equations ◆<br>Please help !

Mila [183]

I'm sure there's an easier way of solving it than the way I did, but I'm not sure what it could be. Never dealt with a problem like this before.

Anyway, I just plugged in and tested. Chose random values for a, b, c, and d, which follow the rule 0 < a < b < c < d:

a = 1
b = 2
c = 3
d = 4

$\sf ax^2+(1-a(b+c))x+abc-d)$

$\sf 1x^2+(1-1(2+3))x+(1)(2)(3)-(4))$

Simplify into standard form:

$\sf x^2+(1-1(5))x+6-4$

$\sf x^2+(1-5)x+2$

$\sf x^2-4x+2$

Use the quadratic formula to solve:

$\sf x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

For functions in the form of $\sf ax^2+bx+c$ . So in this case:

a = 1
b = -4
c = 2

Plug them in:

$\sf x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$

Solve for 'x':

$\sf x=\dfrac{4\pm\sqrt{16-8}}{2}$

$\sf x=\dfrac{4\pm\sqrt{8}}{2}$

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$\sf x\approx 0.59,3.41$

So the answer would be A.

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3 years ago

Theresa Morgan, a chemist, has a 30% hydrochloric acid solution and a 50% hydrochloric acid solution. How many liters of each sh

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Let x represent the number of liters of 50% acid Theresa puts into the mix. The the number of liters of 30% acid will be (420-x). The total amount of acid in the final solution will be ...
0.50x + 0.30(420-x) = 0.45(420)
0.20x + 126 = 189 . . . . . . . . . . . . . . . simplify
0.20x = 63 . . . . . . . . . . . . . . . . . . . . . subtract 126
x = 63/0.20 = 315 . . . . . . . . . . . . . . . liters of 50% solution
(420-x) = 420-315 = 105 . . . . . . . . . liters of 30% solution

Theresa should mix ...
105 liters of 30% solution
315 liters of 50% solution

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3 years ago

points a,b and c are collinear on AC, and AB:BC=3/4. A is located at (x,y), B is located at (4,1), and c is located at (12,5). w

olga_2 [115]

We have two unknowns: x and y. Now, we have to formulate 2 equations. The first would come from the use of the given ratio:

We use the distance formula to find the distance between coordinates:

3/4 = √[(x-4)²+(y-1)²] / √[(4-12)²+(1-5)²]
√[(x-4)²+(y-1)²] = 3√5
(x-4)²+(y-1)² = 45
x² - 8x + 16 + y² - 2y + 1 = 45
x² - 8x + y² - 2y = 28 --> eqn 1

The second equation must come from the equation of a line:
y = mx +b
m = (5-1)/(12-4) = 1/2

Substitute y=5 and x=12 for point (12,5)
5 = (1/2)(12) + b
b = -1

So, the second equation is

y = 1/2x -1 or x = 2 + 2y --> eqn 2

Solving the equations simultaneously:
(2 + 2y)² - 8(2 + 2y) + y² - 2y = 28
Solving for y,
y = -2
x = 2+2(-2) = -2

Therefore, the coordinates of point A is (-2,-2).

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