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Tasya [4]
3 years ago
14

What is x/2=12? I need it for my homework thank you

Mathematics
1 answer:
sveta [45]3 years ago
3 0
X/2 = 12
X = 12 * 2
X = 24
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Which expression represents the distance between point (0, a) and point (a, 0) on a coordinate grid?
o-na [289]

Answer:

\sqrt{2} a

Step-by-step explanation:

The distance between point (x1,y1) and (x2,y2) on a coordinate plane is given by expression \sqrt{(x1-x2)^2+(y1-y2)^2}

In the problem given coordinate points are

(0, a) and (a, 0).

Thus, distance between these points are given by

distance = \sqrt{(x1-x2)^2+(y1-y2)^2}\\=>distance = \sqrt{(0-a)^2+(a-0)^2}\\=>distance = \sqrt{(a)^2+(a)^2}\\\\=>distance = \sqrt{2a^2}\\=>distance = \sqrt{2} a

Thus , expression \sqrt{2} a gives the distance between point (0, a) and point (a, 0) on a coordinate grid.

5 0
3 years ago
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Law of cosines: a2 = b2 + c2 – 2bccos(A) Which equation correctly applies the law of cosines to solve for an unknown angle measu
Cloud [144]
The law of cosines is used for calculating one side of the triangle if the angle opposite to that side is given as well as the other sides. For this problem, we are given angle B. Therefore, the correct answer among the choices given is option 1 where a is equal to 5 and c is equal to 3.
8 0
3 years ago
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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
10 Points! Help is aprreciated!! :)
Ugo [173]

Answer:

A

Step-by-step explanation:

Since the the value is greater than 1, f(x),g(x), and h(x) are exponential functions.

4 0
3 years ago
Can you guys help me with this?
Ghella [55]
F(x)=2x^2-x-6
Factoring:
f(x)=2(2x^2-x-6)/2=(2^2x^2-2x-12)/2=[(2x)^2-(2x)-12]/2
f(x)=(2x-4)(2x+3)/2=(2x/2-4/2)(2x+3)→f(x)=(x-2)(2x+3)

g(x)=x^2-4
Factoring
g(x)=[sqrt(x^2)-sqrt(4)][sqrt(x^2)+sqrt(4)]
g(x)=(x-2)(x+2)

f(x)/g(x)=[(x-2)(2x+3)] / [(x-2)(x+2)
Simplifying:
f(x)/g(x)=(2x+3)/(x+2)

Answer: Third Option (2x+3)/(x+2)
7 0
3 years ago
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