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klio [65]
4 years ago
13

Express the following rate as a rate.

Mathematics
2 answers:
mixas84 [53]4 years ago
6 0
So u do 63/3 so that is 21
bonufazy [111]4 years ago
6 0

Answer: 21 Jumping jacks per minute.


Step-by-step explanation:  63 Divided by 3


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A bus can hold up to 30 students. 10 have already got in class. the inequality is<br>​
stiv31 [10]

Answer:

20? i dont know hopefully im right

Step-by-step explanation:

3 0
3 years ago
Solve the exponential equation<br> 16 = 2^x − 2.
Sholpan [36]
2^x-2=16
+2 +2
——————
2^x=18

(Log)(2))=log (18)

X=log(18)
————-
Log(2)

X=4.169925
3 0
3 years ago
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Darius is a broker who earns a $68,000 annual salary, a 3.15% commission on his clients’ investments, and a fee of $4.25 for eac
taurus [48]

Answer:

$187,244

Step-by-step explanation:

$68,000 annual salary + 3.15% commission + fee of $4.25 per transaction

$68,000 + 0.0315 * $3,600,000 + $4.25 * 1,375 =

= $187,243.75

Answer: $187,244

4 0
3 years ago
Let p= x^2-7.
Advocard [28]

Since p = x² - 7, you can substitute/plug in p for x² - 7

So:

(x² - 7)² - 4x² + 28 = 5

(p)² - 4x² + 28 = 5      You can factor out -4 from (-4x² + 28)

p² - 4(x² - 7) = 5        Plug in p

p² - 4p = 5        Subtract 5

p² - 4p - 5 = 0        Your answer is C

8 0
3 years ago
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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