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viva [34]
3 years ago
11

If there is 20 students then 30 students what is the percentage?

Mathematics
1 answer:
12345 [234]3 years ago
4 0

Answer:

what

Step-by-step explanation:

???

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Help please and please give a detailed answer
Natali [406]
The answer is 18
Working out:
1 1/2 x 4 =6
6 x 3=18
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Evaluate 5(14 – 23) + 8 ÷ 2. A. 26 B. 34 C. 44 D. 56
Keith_Richards [23]

Answer:

We multiplu5(14-23)which will vive us 115-70 equivalent to -45 and then we divide 8÷2which will give us 4. Finally we add -45+4=-41

Step-by-step explanation:

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Pls help fast its urgent
telo118 [61]

Answer: -8

Step-by-step explanation:

3 - (-8) = 3+8

3+ 8=11

6 0
3 years ago
if 1/4 of a gallon of paint is needed to paint 2/3 of a fence, how many gallons are needed to paint the entire fence? site:socra
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\bf \begin{array}{ccll} \stackrel{paint}{gallons}&fence\\ \cline{1-2} \\\frac{1}{4}&\frac{2}{3}\\\\ x&1 \end{array}\implies \cfrac{~~\frac{1}{4}~~}{x}=\cfrac{~~\frac{2}{3}~~}{1}\implies \cfrac{~~\frac{1}{4}~~}{\frac{x}{1}}=\cfrac{2}{3}\implies \cfrac{1}{4}\cdot \cfrac{1}{x}=\cfrac{2}{3} \\\\\\ \cfrac{1}{4x}=\cfrac{2}{3}\implies 3=8x\implies \cfrac{3}{8}=x

7 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
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