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3 years ago

P and Q are points on the line 3y - 4x = 12

Mathematics

1 answer:

Lorico [155]3 years ago

6 0

Answer:

Step-by-step explanation:

Since the coordinates of P are (0, 1), this makes P the y-intercept of that line. The y-intercept exists where x = 0. And in the coordinate (0, 1), x does in fact equal 0.

Since the one coordinate given in Q is (?, 0), this means that Q is the x-intercept of the line. The x-intercept exists where y = 0. And in the coordinate (?, 0), y does in fact equal 0. So in order to solve for the x coordinate of Q, we plug in a 0 for y and solve for x:

3(0) - 4x = 12 and

-4x = 12 so

x = -3

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3 years ago

Algebra problem, please help

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3 years ago

Read 2 more answers

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4

vlabodo [156]

Answer:

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

Step-by-step explanation:

Bernoulli's equation,

$P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2$

P₁ = P₂= atmospheric presser

$\rho$ = density

$\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2$ [since P₁ = P₂]

$\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)$

$\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2$

$\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2$

$\Rightarrow v^2_2- v^2_1=2g h$ [ $h_1-h_2=h$ ]

Here $v_1\approx 0$

$\Rightarrow v^2_2=2g h$

$\therefore v_2=\sqrt {2gh$

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole $v_2$ = $\sqrt {2\times 9.8\times 10.4}$ m/s

=14.28 m/s.

Given, the rate of flow from the leak is $2.53\times 10^{-3}$ $m^3/min$

$=\frac{2.53\times 10^{-3}}{60}$ $m^3/s$

Let the diameter of the hole be d.

Then the cross section area of the hole is $=\pi (\frac d2)^2$

We know that,

The rate of flow = Cross section area × speed

$\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28$

$\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}$

$\Rightarrow d= 1.94 \times 10^{-3}$

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

4 0

4 years ago

A model rocket is launched with an initial upward velocity of 57m/s. The rocket's height h (in meters) after t seconds is given

vesna_86 [32]

The values of for which the rocket's height is 29 meters are 0.75 and -0.67.

<h3>What is a velocity?</h3>

Velocity can be described as a vector quantity having both direction and magnitude which is the distance traveled per time.

We were given the rocket's height h as $h=57t-5t^2$ which is an equation, and we were been given the height which is the value that can be input into the equation as is 29 meters , then we can proceed to simplify the equation as

$h=57t-5t^2$

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then we will have $57t-5t^2-29=0$

with the use of quadratic formula the values of t are : 0.75 and -0.67

Then we can chose 0.75 seconds

Therefore, after the calculation with the use of the quadratic equation, we can come into conclusion that the values that are required in the question are 0.75 and -0.67.

Learn more about velocity at:

brainly.com/question/4535155

#SPJ1

NOTE: This is a complete question, there were no given options, even on the internet.

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1 year ago

Easy points! 35 points and brainliest

emmainna [20.7K]

Answer:

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