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Vikki [24]
3 years ago
12

A. y = -1 B. X = -1 C. y = 1 D. X = 1

Mathematics
1 answer:
Marizza181 [45]3 years ago
3 0
Answer- D X=1
The highest point in the number is X=1
You might be interested in
1-What is the sum of the series? ​∑j=152j​ Enter your answer in the box.
tangare [24]

Answer:

Please see the Step-by-step explanation for the answers

Step-by-step explanation:

1)

∑\left \ {{5} \atop {j=1}} \right. 2j

The sum of series from j=1 to j=5 is:

∑ = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)

  =  2 + 4 + 6 + 8 + 10

∑ = 30

2)

This question is not given clearly so i assume the following series that will give you an idea how to solve this:

∑\left \ {{4} \atop {k=1}} \right. 2k²

The sum of series from k=1 to j=4 is:

∑ = 2(1)² + 2(2)² + 2(3)² + 2(4)²

  = 2(1) + 2(4) + 2(9) + 2(16)

  =  2 + 8 + 18 + 32

∑ = 60

∑\left \ {{4} \atop {k=1}} \right. (2k)²

∑ = (2*1)² + (2*2)² + (2*3)² + (2*4)²

  = (2)² + (4)² + (6)² + (8)²

  = 4 + 16 + 36 + 64

∑ = 120

∑\left \ {{4} \atop {k=1}} \right. (2k)²- 4

∑ = (2*1)²-4 + (2*2)²-4 + (2*3)²-4 + (2*4)²-4

  = (2)²-4 + (4)²-4 + (6)²-4 + (8)²-4

  = (4-4) + (16-4) + (36-4) + (64-4)

  = 0 + 12 + 32 + 60

∑ = 104

∑\left \ {{4} \atop {k=1}} \right. 2k²- 4

∑ = 2(1)²-4 + 2(2)²-4 + 2(3)²-4 + 2(4)²-4

  = 2(1)-4 + 2(4)-4 + 2(9)-4 + 2(16)-4

  = (2-4) + (8-4) + (18-4) + (32-4)

  = -2 + 4 + 14 + 28

∑ = 44

3)

∑\left \ {{6} \atop {k=3}} \right. (2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)  

  = (6-10) + (8-10) + (10-10) + (12-10)

  = -4 + -2 + 0 + 2  

∑ = -4

4)

1+1/2+1/4+1/8+1/16+1/32+1/64

This is a geometric sequence where first term is 1 and the common ratio is 1/2 So

a = 1

This can be derived as

1/2/1 = 1/2 * 1 = 1/2

1/4/1/2 = 1/4 * 2/1 = 1/2

1/8/1/4 = 1/8 * 4/1  = 1/2

1/16/1/8 = 1/16 * 8/1  = 1/2

1/32/1/16 = 1/32 * 16/1  = 1/2

1/64/1/32 = 1/64 * 32/1  = 1/2

Hence the common ratio is r = 1/2

So n-th term is:

ar^{n-1} = 1(\frac{1}{2})^{n-1}

So the answer that represents the series in sigma notation is:

∑\left \ {{7} \atop {j=1}} \right. (\frac{1}{2})^{j-1}

5)

−3+(−1)+1+3+5

This is an arithmetic sequence where the first term is -3 and the common difference is 2. So  

a = 1

This can be derived as

-1 - (-3) = -1 + 3 = 2

1 - (-1) = 1 + 1 = 2

3 - 1 = 2

5 - 3 = 2

Hence the common difference d = 2

The nth term is:

a + (n - 1) d

= -3 + (n−1)2

= -3 + 2(n−1)

= -3 + 2n - 2

= 2n - 5

So the answer that represents the series in sigma notation is:

∑\left \ {{5} \atop {j=1}} \right. (2j−5)

6 0
3 years ago
The graph shows the data points in the table and the exponential regression model associated with the data ?
Ipatiy [6.2K]
From the graph, it is obvious that the trend is decreasing from 100 on day 2, to 1 on day 10. So, the answer could either be A or C. The question would be how fast is it decreasing? To illustrate this, let's find the difference of consecutive data:

100 - 26 = 74
26 - 6 = 20
6 - 2=4
2-1=1

It must not be an additive rate because there is no common difference. Let's illustrate if the trend is in multiplicative rate:

100/26 = 3.85
26/6 = 4.33
6/2 = 3
2/1 = 2

More or less, they have a common divider. Hence, the decreasing rate is in multiplicative rate. The answer is A.
6 0
3 years ago
Read 2 more answers
Help please! Not sure what this is
Lesechka [4]
B. Rotation
I’m assuming the question is not asking about placement but just the way the shape is.
Hope this helps!
3 0
3 years ago
Read 2 more answers
Julio signs a new lease for $800 and in the agreement the complex will raise his rent 3% each year that he renews. As of right n
Sholpan [36]
D because it does not specify when he would buy a house
7 0
3 years ago
b) If parametric equations of a flow line are x = x(t), y = y(t), explain why these functions satisfy the differential equations
sineoko [7]

Answer:

The equation of the the flow line that passes through the point (x, y) = (−1, −1) is

In y + In x = 0 or in another form, xy = 1.

Step-by-step explanation:

The pathline equation for a vector field is given by F(x,y) = xî - yj

The velocity vector field for the streamline of the flow is given by

V(x, y) = (dx/dt)î + (dy/dt)j

From the question, it is given that

(dx/dt) = x

(dy/dt) = -y

Hence, the velocity vector field for the streamline of the flow in question is

V(x, y) = xî - yj

which coincides with the pathline vector field of the flow.

The only time the pathline and streamline vector field coincide and have the same equation is when the flow is a steady state flow.

That is, the properties of the fluid flowing isn't changing with time!

Hence, this flow is a steady state flow!

We're told to solve the differential equation.

(dx/dt) = x

(dy/dt) = -y

but

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = -y/x

(dy/y) = -(dx/x)

∫(dy/y) = -∫ (dx/x)

In y = - In x + c

where c is the constant of integration

In y + In x = c

In (xy) = c

Inserting the values of (x, y) given in the question,

In (-1 × -1) = c

In 1 = c

0 = c

c = 0

In y + In x = 0

In (yx) = 0

xy = e⁰ = 1

xy = 1

So, the equation of the the flow line that passes through the point (x, y) = (−1, −1) is

In y + In x = 0 or in another form, xy = 1

Hope this Helps!!!

4 0
3 years ago
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