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Arada [10]
3 years ago
14

What is the point slope form of a line that has a slope of 3 and passes through the point (-1,4)

Mathematics
1 answer:
lukranit [14]3 years ago
6 0

Answer:

y - 4 = 3(x- (-1)     --->      y-4 = 3(x+1)

Step-by-step explanation:

y - 4 = 3(x + 1)

y - 4 = 3x + 3

y = 3x + 7

then plug in (-1,4)

4 = 3(-1) + 7

4 = -3 + 7

4 = 4

Point slope form looks like:

y - y₁ = m(x - x₁)

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Use these functions to answer the questions.
Elenna [48]
Part A
We have f\left(x\right)=\left(x+3\right)^2-1. To solve for the x-intercept, we set f(x) equal to 0.  That is
     \left(x+3\right)^2-1=0

     \left(x+3\right)^2=1

Take the square root of both sides, 
     x+3=1

     x=-2

The x-intercept is (-2,0). 

To solve for the y-intercept, we set x=0. That is 
     y=\left(0+3\right)^2-1=3^2-1=9-1=8

The y-intercept is (0, 8) 

The coordinates of the optimum point are actually the vertex which can be easily seen from the vertex form equation given above. The minimum point is (-3, -1).

Part B.
We have g\left(x\right)=-2x^2+8x+3.
Factor out -2
     =-2\left(x^2-4x\right)+3

Complete the square
     =-2\left(x^2-4x+4\right)+3-2\left(4\right)

Simplify
     g(x)=-2\left(x-2\right)^2-5

Part C
We have h\left(t\right)=-16t^2+28t.
The maximum height is 12.25 feet after 0.875 seconds from the time of the jump. The dolphin will be back in the water after 1.75 seconds. The graph of the jump is shown in the photo. 
     
    

4 0
3 years ago
Find the equivalent fraction of 2/3 having denominator 6 and 6 and numerator 10.​
4vir4ik [10]

Step-by-step explanation:

having denominator 6,

X/6= 2/3

X= 4

so, 4/6= 2/3

having numerator 10,

10/X= 2/3

X= 15

so, 10/15= 2/3

7 0
3 years ago
The biggest zucchini from Joe's farm is 2 pounds, which is 1
Vera_Pavlovna [14]

Step-by-step explanation:

It means average weight of zucchinis is 1 pound because 2 - 1 = 1

4 0
3 years ago
EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 11z on the curve of intersection of the plane x − y + z =
Taya2010 [7]

Answer:

\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}

<em>Maximum value of f=2.41</em>

Step-by-step explanation:

<u>Lagrange Multipliers</u>

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

\bigtriangledown  f=\lambda \bigtriangledown  g

for some scalar \lambda called the Lagrange multiplier.

For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is

\bigtriangledown  f=\lambda \bigtriangledown  g+\mu \bigtriangledown  h

The gradient of f is

\bigtriangledown  f=

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in x,y,z,\lambda,\mu.

We have

f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0

Let's compute the partial derivatives

f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0

The Lagrange condition leads to

1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)

Operating and simplifying

1=\lambda+2x\mu\\2=-\lambda +2y\mu \\\lambda=11

Replacing the value of \lambda in the two first equations, we get

1=11+2x\mu\\2=-11 +2y\mu

From the first equation

\displaystyle 2\mu=\frac{-10}{x}

Replacing into the second

\displaystyle 13=y\frac{-10}{x}

Or, equivalently

13x=-10y

Squaring

169x^2=100y^2

To solve, we use the restriction h

x^2 + y^2 = 1

Multiplying by 100

100x^2 + 100y^2 = 100

Replacing the above condition

100x^2 + 169x^2 = 100

Solving for x

\displaystyle x=\pm \frac{10}{\sqrt{269}}

We compute the values of y by solving

13x=-10y

\displaystyle y=-\frac{13x}{10}

For

\displaystyle x= \frac{10}{\sqrt{269}}

\displaystyle y= -\frac{13}{\sqrt{269}}

And for

\displaystyle x= -\frac{10}{\sqrt{269}}

\displaystyle y= \frac{13}{\sqrt{269}}

Finally, we get z using the other restriction

x - y + z = 1

Or:

z = 1-x+y

The first solution yields to

\displaystyle z = 1-\frac{10}{\sqrt{269}}-\frac{13}{\sqrt{269}}

\displaystyle z = \frac{-23\sqrt{269}+269}{269}

And the second solution gives us

\displaystyle z = 1+\frac{10}{\sqrt{269}}+\frac{13}{\sqrt{269}}

\displaystyle z = \frac{23\sqrt{269}+269}{269}

Complete first solution:

\displaystyle x= \frac{10}{\sqrt{269}}\\\\\displaystyle y= -\frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{-23\sqrt{269}+269}{269}

Replacing into f, we get

f(x,y,z)=-0.4

Complete second solution:

\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}

Replacing into f, we get

f(x,y,z)=2.4

The second solution maximizes f to 2.4

5 0
3 years ago
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Novosadov [1.4K]

Answer:

1. A

2. C

Step-by-step explanation:

6 0
3 years ago
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