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Anarel [89]
2 years ago
12

What does 1/5 + (-1/2) equal

Mathematics
2 answers:
KonstantinChe [14]2 years ago
8 0

Answer:

-3/10

Step-by-step explanation:

Ket [755]2 years ago
3 0

Answer: (:

Step-by-step explanation:

-0.3

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Write the phrase as an expression.<br><br> the product of 3+12<br><br> an expression is _____
Dima020 [189]

Answer:

Writing expressions | Math (article) | Khan Academy

Step-by-step explanation:

8 0
3 years ago
What is 7 4/10 -1 1/3
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The answer is 6 and 1/15.
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Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or
NARA [144]

Answer:

Saddle point: (0,0)

Local minimum: (\frac{3}{8}, -\frac{3}{8})

Local maxima: (0,-\frac{9}{8}), (\frac{9}{8},0)

Step-by-step explanation:

The function is:

f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y

The partial derivatives of the function are included below:

\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y

\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)

\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x

\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)

Local minima, local maxima and saddle points are determined by equalizing  both partial derivatives to zero.

y \cdot (8\cdot y -16\cdot x + 9) = 0

x \cdot (16\cdot y - 8\cdot x + 9) = 0

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

x\cdot (-8\cdot x + 9) = 0

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

y\cdot (8\cdot y+9) = 0

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}

The second derivatives of the function are:

\frac{\partial^{2} f}{\partial x^{2}} = 0

\frac{\partial^{2} f}{\partial y^{2}} = 0

\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9

Then, the expression is simplified to this and each point is tested:

H = -16\cdot y +16\cdot x -9

S1: (0,0)

H = -9 (Saddle Point)

S2: (3/8,-3/8)

H = 3 (Local maximum or minimum)

S3: (9/8, 0)

H = 9 (Local maximum or minimum)

S4: (0, - 9/8)

H = 9 (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64} (Local minimum)

S3: (9/8, 0)

f(\frac{9}{8},0) = 0 (Local maximum)

S4: (0, - 9/8)

f(0,-\frac{9}{8} ) = 0 (Local maximum)

Saddle point: (0,0)

Local minimum: (\frac{3}{8}, -\frac{3}{8})

Local maxima: (0,-\frac{9}{8}), (\frac{9}{8},0)

4 0
3 years ago
AABC is a translation of AABC. What is the length of AC'?
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Would appreciate some help with this:<br><br> Verify (sinq+cosq)^2 - 1 = sin2q
Phantasy [73]

Hello from MrBillDoesMath!

Answer:

See Discussion below

Discussion:

(sinq + cosq)^2 =                                     => (a +b)^2 = a^2 + 2ab + b^2

(sinq)^2 + (cosq)^2 + 2 sinq* cosq        => as (sinx)^2 + (cosx)^2 = 1

1 + 2 sinq*cosq              (*)


Setting a = b = q in the trig identity:

sin(a+b) = sina*cosb + cosa*sinb                      

sin(2q) =                      (**)

sinq*cosq + cosq*sinq      => as both terms are identical

2 sinq*cosq


Combining (*) and (**)

(sinq + cosq)^2 = 1 + 2sinq*cosq     => (**) 2sinq*cosq =  sqin(2q)

                          = 1 +  sin(2q)

Hence

(sinq + cosq)^2  = 1 +  sin(2q)            => subtracting 1 from both sides

(sinq + cosq)^2  - 1 =   sin(2q)  

The last statement is what we are trying to prove.



Thank you,

MrB

7 0
3 years ago
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