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VMariaS [17]
3 years ago
15

Mikayla withdraws $20 on 4 different days durin the week.Find the total change in her account balance after the withdrawls.

Mathematics
1 answer:
Stells [14]3 years ago
7 0

Wait what did she have in her account? We need to know to answer.

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How to do this. Can you maybe show me your work?
Musya8 [376]
VX = 204
Divide by 3 to get the 2/3 , 1/3 ratio of the segments
204/3= 68
68*2= 136
VW= 136, XW= 68
Do the same for RW, RY
RW= 104 this is 2/3 of the segment. Divide by 2.
WY= 52. RY= 156

#7
Find the midpoint of each side. (0,2) (7,4) m1=(3.5,3)
(0,3) (5,0) m2=(2.5,1.5)
(5,0) (7,4) m3=(6,2)
Draw a segment from each midpoint to its opposite vertex. The point of intersection is (4,2)
7 0
3 years ago
A prism with a base area of 3 m² and a height of 4 m is dilated by a factor of 3/2 . What is the volume of the dilated prism?
erma4kov [3.2K]
Volume of Original Prism = base area * Height = 3 * 4 = 12

After dilation factor = 12 * 3/2 = 6 * 3 = 18

In short, Your Answer would be 18 m³

Hope this helps!
4 0
3 years ago
Read 2 more answers
IF YOU GET ANSWR RIGHT YOU GET BRAINLIEST!
ollegr [7]

Answer:

b

Step-by-step explanation:

4 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
Express 30 as a product of its prime factors
Olegator [25]

Answer:

5 * 3 * 2

Step-by-step explanation:

30

15 * 2

5 * 3 * 2

3 0
2 years ago
Read 2 more answers
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