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3 years ago

PLS HELP.....

Mathematics

1 answer:

Gemiola [76]3 years ago

3 0

The second one because Idk I’m just guessing lol it doesn’t look like it should be there though

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Solve for x. Enter the solutions from least to greatest.

Anna [14]

Answer:

lesser x = -6

greater x = 12

Step-by-step explanation:

(x - 3)^2 - 81 = 0

(x - 3)^2 = 81

x - 3 = +/- 9

x - 3 = 9 or x - 3 = -9

x = 12 or x = -6

5 0

3 years ago

PLS TRY TO ANSWER ASAP.

Harrizon [31]

Answer:

0.075 m/min

Explanation:

You need to use derivatives which is an advanced concept used in calculus.

1. Write the equation for the volume of the cone:

$V=\dfrac{1}{3}\pi r^2h$

2. Find the relation between the radius and the height:

r = diameter/2 = 5m/2 = 2.5m

h = 5.2m

h/r =5.2 / 2.5 = 2.08

3. Filling the tank:

Call y the height of water and x the horizontal distance from the axis of symmetry of the cone to the wall for the surface of water, when the cone is being filled.

The ratio x/y is the same r/h

x/y=r/h

y = x . h / r

The volume of water inside the cone is:

$V=\dfrac{1}{3}\pi x^2y$

$V=\dfrac{1}{3}\pi x^2(2.08)\cdot x\\\\\\V=\dfrac{2.08}{3}\pi x^3$

4. Find the derivative of the volume of water with respect to time:

$\dfrac{dV}{dt}=2.08\pi x^2\dfrac{dx}{dt}$

5. Find x² when the volume of water is 8π m³:

$V=\dfrac{2.08}{3}\pi x^3\\\\\\8\pi=\dfrac{2.08}{3}\pi x^3\\\\\\ 11.53846=x^3\\ \\ \\ x=2.25969\\ \\ \\ x^2=5.1062$ m²

6. Solve for dx/dt:

$1.2m^3/min=2.08\pi(5.1062m^2)\dfrac{dx}{dt}$

$\dfrac{dx}{dt}=0.03596m/min$

7. Find dh/dt:

From y/x = h/r = 2.08:

$y=2.08x\\\\\\\dfrac{dy}{dx}=2.08\dfrac{dx}{dt}\\\\\\\dfrac{dy}{dt}=2.08(0.035964m/min)=0.0748m/min\approx0.075m/min$

That is the rate at which the water level is rising when there is 8π m³ of water.

4 0

3 years ago

The rule for the nth term of a sequence is 5n + 16 Find the 10th term of the sequence.

lara [203]

Answer:

Step-by-step explanation:

It's given that

Tn = 5n + 16

The 10th term is

T10 = 5(10) + 16

= 66

8 0

2 years ago

Read 2 more answers

Assign point_dist with the distance between point (x1, y1) and point (x2, y2). the calculation is: distance = squarerootof( (x2

MissTica

import math def calculateDistance(x1,y1,x2,y2): dist = math.sqrt((x2 - x1)**2 + (y2 - y1)**2) return dist distance = calculateDistance(2,4,6,8) print distance

8 0

3 years ago

PLEASE HELP ALGEBRA 2!!!

BlackZzzverrR [31]

Step-by-step explanation:

-4.5x -2y = -12.5 -------(1)

3.25x -y = -0.75 --------(2)

From (2) we can make y the subject of the relation

y = 0.75+3.25x -------(3)

Input 0.75+3.25x into y in (1)

-4.5x -2(0.75+3.25x) = -12.5

-4.5x -1.5- 6.5x = -12.5

-4.5x -6.5x -1.5 = -12.5

-11x = -12.5 +1.5

-11x = -11

x = 11/11

x = 1

If x = 1,substitute 1 into x in (3)

y = 0.75+3.25x

y = 0.75 +3.25(1)

y = 0.75 +3.25

y = 4

x = 1,y = 4

7 0

3 years ago