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Lyrx [107]
4 years ago
8

Help another surface area here!!!

Mathematics
1 answer:
son4ous [18]4 years ago
6 0
The surface area is 14.0095846890 I think
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What is the slope of the line that passes through the points (-6,1) and (4,-4)
mr Goodwill [35]

Answer:

-1/2

Step-by-step explanation:

change in y/ change in x = -4-1/4-(-6) =-5/10 = -1/2

4 0
3 years ago
If a polynomial function f(x) has roots 4 – 13i and 5, what must be a factor of f(x)? (x (13 – 4i)) (x – (13 4i)) (x (4 13i)) (x
Fiesta28 [93]

The polynomial function f(x) with roots 4 – 13i and 5 have factor of x - (4 - 13)

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more number and variables.

Polynomial function f(x) has roots 4 – 13i and 5. Hence:

x = 4 - 13i; and x = 5

x - (4 - 13) = 0; and x - 5 = 0

f(x) = [x - (4 - 13)](x - 5)

The polynomial function f(x) with roots 4 – 13i and 5 have factor of x - (4 - 13)

Find out more on equation at: brainly.com/question/2972832

#SPJ4

5 0
2 years ago
Can someone help me with this please?
Elena-2011 [213]
29.) This is simply interpreting the linear function.
The function:
96+2.1x
a.) The 96 is what you start with from the beginning. So, we know Rachel weighed 96 ounces when she was born.
b.) Again, interpretation. The 2.1 is the slope, or the change in the line. So we know that Rachel gains 2.1 ounces each day.
30.) For this one, you just have to plug 9 into the equation for x.
6 0
3 years ago
The membership fee for joining a camp association is $45 a local campground charges members of the camping association $35 per n
sleet_krkn [62]
45 + 35x = 40x
45 = 40x - 35x
45 = 5x
45/5 = x
9 = x <=== they are the same cost at 9 nights
8 0
4 years ago
Find the minimum and maximum of f(x,y,z)=x2+y2+z2 subject to two constraints, x+2y+z=7 and x−y=6.
sweet-ann [11.9K]
Use the method of Lagrange multipliers. We have Lagrangian

L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-7)+\lambda_2(x-y-6)

with partial derivatives (set equal to 0) of

L_x=2x+\lambda_1+\lambda_2=0
L_y=2y+2\lambda_1-\lambda_2=0
L_z=2z+\lambda_1=0
L_{\lambda_1}=x+2y+z-7=0
L_{\lambda_2}=x-y-6=0

As x+2y+z=7, and x-y=6, we can obtain

\dfrac12L_x+L_y+\dfrac12L_z=0\implies3\lambda_1-\dfrac12\lambda_2=-7
L_x-L_y=0\implies\lambda_1-2\lambda_2=12
\begin{cases}3\lambda_1-\frac12\lambda_2=-7\\\lambda_1-2\lambda_2=12\end{cases}\implies\lambda_1=-\dfrac{40}{11},\lambda_2=-\dfrac{86}{11}

From this, we find a single critical point:

2x-\dfrac{40}{11}-\dfrac{86}{11}=0\implies x=\dfrac{63}{11}
\dfrac{63}{11}-y=6\implies y=-\dfrac3{11}
\dfrac{63}{11}-\dfrac6{11}+z=7\implies z=\dfrac{20}{11}

At this point, we have a value of

f\left(\dfrac{63}{11},-\dfrac3{11},\dfrac{20}{11}\right)=\dfrac{398}{11}

To determine what kind of extremum occurs at this point, we check the Hessian of f(x,y,z):

\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}

We observe that \det\mathbf H(x,y,z)=8>0 at any point (x,y,z), and that the eigenvalues of this matrix are all positive (2 with multiplicity 3), so \mathbf H is positive definite. By the second partial derivative test, this means f(x,y,z) attains a minimum at this critical point. Meanwhile, f has no maximum value.
5 0
3 years ago
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