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krek1111 [17]
3 years ago
5

Miss Watson runs a distance of 200 meters in 25 seconds

Mathematics
1 answer:
gogolik [260]3 years ago
4 0

Answer:

8m/s

Step-by-step explanation:

Average speed = distance/time taken

So

200m/25s

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100-99+98-97+96-95+...+8-7+6-5+4-3+2-1
lina2011 [118]
100-99+98-97+96-95+...+8-7+6-5+4-3+2-1\\\\=(100-99)+(98-97)+(96-95)+...+(4-3)+(2-1)\\\\=\underbrace{1+1+1+...+1}_{50}=\fbox{50}

8 0
3 years ago
Read 2 more answers
A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station. Find the ra
Anika [276]

Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

Step-by-step explanation:

Given information:

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.

z=1

\frac{dx}{dt}=430

We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

y=2

According to Pythagoras

hypotenuse^2=base^2+perpendicular^2

y^2=x^2+1^2

y^2=x^2+1               .... (1)

Put z=1 and y=2, to find the value of x.

2^2=x^2+1^2

4=x^2+1

4-1=x^2

3=x^2

Taking square root both sides.

\sqrt{3}=x

Differentiate equation (1) with respect to t.

2y\frac{dy}{dt}=2x\frac{dx}{dt}+0

Divide both sides by 2.

y\frac{dy}{dt}=x\frac{dx}{dt}

Put x=\sqrt{3}, y=2, \frac{dx}{dt}=430 in the above equation.

2\frac{dy}{dt}=\sqrt{3}(430)

Divide both sides by 2.

\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}

\frac{dy}{dt}=372.390923627

\frac{dy}{dt}\approx 372

Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

6 0
3 years ago
The deepest part of the Mariana trench (the deepest trench in the world’s oceans) is at an elevation of -11.033 kilometers. The
SVETLANKA909090 [29]

Answer:

Step-by-step explanation:

-2.968

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
What us the intercept form of (1,2) and (4,-4)
lutik1710 [3]

first you need to find the slope (m) using y2-y1/x2-x1.

-4-2/4-1= -6/3=-2

m=-2

then plug into the intercept form formula: y-y1=m(x-x1)

y-2=-2(x-1)

5 0
2 years ago
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