Came

PLEASE HELP!!!!!! Find the image of (1,2) after a reflection about x=6 followed by a reflection about x=4

Bond [772]

I would help you but it me hard for me too

5 0

3 years ago

Find a point on the curve x^3+y^3=11xy other than the origin at which the tangent line is horizontal.

attashe74 [19]

Compute the derivative dy/dx using the power, product, and chain rules. Given

x³ + y³ = 11xy

differentiate both sides with respect to x to get

3x² + 3y² dy/dx = 11y + 11x dy/dx

Solve for dy/dx :

(3y² - 11x) dy/dx = 11y - 3x²

dy/dx = (11y - 3x²)/(3y² - 11x)

The tangent line to the curve is horizontal when the slope dy/dx = 0; this happens when

11y - 3x² = 0

or

y = 3/11 x²

(provided that 3y² - 11x ≠ 0)

Substitute y into into the original equation:

x³ + (3/11 x²)³ = 11x (3/11 x²)

x³ + (3/11)³ x⁶ = 3x³

(3/11)³ x⁶ - 2x³ = 0

x³ ((3/11)³ x³ - 2) = 0

One (actually three) of the solutions is x = 0, which corresponds to the origin (0,0). This leaves us with

(3/11)³ x³ - 2 = 0

(3/11 x)³ - 2 = 0

(3/11 x)³ = 2

3/11 x = ³√2

x = (11•³√2)/3

Solving for y gives

y = 3/11 x²

y = 3/11 ((11•³√2)/3)²

y = (11•³√4)/3

So the only other point where the tangent line is horizontal is ((11•³√2)/3, (11•³√4)/3).

3 0

2 years ago

Hurrrryyyyy plllzzz. Determine the equation of the line that passes through the given points

pashok25 [27]

<h2>Answer:</h2>

y = 2

<h2>Step-by-step explanation:</h2>

To determine the equation of the line that passes through (10,2) and (-3,2), we need to determine the slope of the line. Then substitute the slope and any given point in point slope form to obtain the equation of the line.

<h3>Finding the Slope of the line:</h3>

$\implies \text{Slope} = \dfrac{\text{Rise}}{\text{Run}} = \dfrac{y_{2} - y_{1} }{x_{2} - x_{1} }$

$\implies \text{Slope} = \dfrac{y_{2} - y_{1} }{x_{2} - x_{1} }$

<u>Substitute the coordinates of the given points:</u>

$\implies \text{Slope} = \dfrac{2 - 2}{-3 - 10 }$

<u>Simplify the equation to determine the slope:</u>

$\implies \text{Slope} = \dfrac{0}{-3 - 10 }$

∴ 0 divided by ANY number is ALWAYS 0.

$\implies \text{Slope} =0$

<h3>Finding the equation of the line:</h3>

Point slope form formula: y - y₁ = m(x - x₁)

x₁ and y₁ are the coordinates of any given point.
m is the slope

<u>Substitute the values in the point slope form:</u>

$\implies \text{Equation of line:} \ y - y_{1} = m(x - x_{1} )$

$\implies \text{Equation of line:} \ y - 2 = 0(x - 10)$

<u>Simplify the equation to determine the equation of the line:</u>

∴ Any number multiplied by 0 is 0.

$\implies \text{Equation of line:} \ y - 2 = 0$

$\implies \text{Equation of line:} \ y = 0 + 2$

$\implies \text{Equation of line:} \ y = 2$

Thus, the equation of the line is y = 2.

5 0

3 years ago

There were 29 students available for the woodwind section of the school orchestra. 11 students could play the flute, 15 could pl

Dafna11 [192]

Answer:

a. The number of students who can play all three instruments = 2 students

b. The number of students who can play only the saxophone is 0

c. The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only one of the clarinet, saxophone, or flute = 4

Step-by-step explanation:

The total number of students available = 29

The number of students that can play flute = 11 students

The number of students that can play clarinet = 15 students

The number of students that can play saxophone = 12 students

The number of students that can play flute and saxophone = 4 students

The number of students that can play flute and clarinet = 4 students

The number of students that can play clarinet and saxophone = 6 students

Let the number of students who could play flute = n(F) = 11

The number of students who could play clarinet = n(C) = 15

The number of students who could play saxophone = n(S) = 12

We have;

a. Total = n(F) + n(C) + n(S) - n(F∩C) - n(F∩S) - n(C∩S) + n(F∩C∩S) + n(non)

Therefore, we have;

29 = 11 + 15 + 12 - 4 - 4 - 6 + n(F∩C∩S) + 3

29 = 24 + n(F∩C∩S) + 3

n(F∩C∩S) = 29 - (24 + 3) = 2

The number of students who can play all = 2

b. The number of students who can play only the saxophone = n(S) - n(F∩S) - n(C∩S) - n(F∩C∩S)

The number of students who can play only the saxophone = 12 - 4 - 6 - 2 = 0

The number of students who can play only the saxophone = 0

c. The number of students who can play the saxophone and the clarinet but not the flute = n(C∩S) - n(F∩C∩S) = 6 - 2 = 4

The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only the saxophone = 0

The number of students who can play only the clarinet = n(C) - n(F∩C) - n(C∩S) - n(F∩C∩S) = 15 - 4 - 6 - 2 = 3

The number of students who can play only the clarinet = 3

The number of students who can play only the flute = n(F) - n(F∩C) - n(F∩S) - n(F∩C∩S) = 11 - 4 - 4 - 2 = 1

The number of students who can play only the flute = 1

Therefore, the number of students who can play only one of the clarinet, saxophone, or flute = 1 + 3 + 0 = 4

The number of students who can play only one of the clarinet, saxophone, or flute = 4.

6 0

4 years ago

6th grade math help me pleaseeee

Furkat [3]

Answer:

I think it's the top left one

Step-by-step explanation:

8 0

3 years ago

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