Apply Runge-Kutta method of second order to find an approximate value of y when x=0.02, for first order initial value problem [10 Marks] y = x² + y, y(0) = 1. Assume step-size (h) as 0.01. Apply Runge-Kutta method of second order to find an approximate value of y when x=0.02, for first order initial value problem y = x² + y, y(0) = 1. Assume step-size (h) as 0.01.

Answer:

first option

Step-by-step explanation:

two solutions means 2 cutting points

Answer: 2x + 6y + 22 units

Step-by-step explanation:

The perimeter of a figure is the total length of all the sides of a figure.

Therefore, the perimeter can be found by adding up all the lengths of the figure.

Suppose that P is the perimeter of this figure, then

[tex]P = x+4+x+4+3y+7+7+3y\\[/tex]

Combining like terms, we get

[tex]P = 2x + 6y + 22[/tex]

Since we do not have any information on what the variables are, this is the perimeter of the figure.

Answer:

15

Step-by-step explanation:

The blue angle and the yellow angle form a linear pair, so they are supplementary (angle measures add up to 180).

180 - 165 = 15

Answer:

Shark

Step-by-step explanation:

They have gills,scales,swim bladders to float,most produce eggs and also in ectothermic.

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Further explanation:

Answer:

the scale factor from triangle ABC to triangle DEF would be 8

Step-by-step explanation:

the triangle DEF was actually flipped from triangle ABC

1.line AB corresponds to line EF

AB = 6 , and EF = 48

we multiply 6 by "8" to get to 48

2.line BC corresponds to line ED

BC = 6 , and ED = 48

we multiply 6 by "8" to get to 48

3.line AC corresponds to line FD

AC = 6 , and FD = 48

we multiply 4 by "8" to get to 32

I hope this help

Answer:-2.4+3.2t

Step-by-step explanation:

-3.6+1.2-1.9t+5.1t

= -2.4+3.2t

The equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2

A quadratic function is represented as:

y = a(x - h)^2 + k

Question #6

The vertex of the graph is

(h, k) = (-1, 2)

So, we have:

y = a(x + 1)^2 + 2

The graph pass through the f(0) = -2

So, we have:

-2 = a(0 + 1)^2 + 2

Evaluate the like terms

a = -4

Substitute a = -4 in y = a(x + 1)^2 + 2

y = -4(x + 1)^2 + 2

Question #7

The vertex of the graph is

(h, k) = (2, 1)

So, we have:

y = a(x - 2)^2 + 1

The graph pass through (1, 3)

So, we have:

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

Evaluate the like terms

a = 2

Substitute a = 2 in y = a(x - 2)^2 + 1

y = 2(x - 2)^2 + 1

Question #8

The vertex of the graph is

(h, k) = (1, -2)

So, we have:

y = a(x - 1)^2 - 2

The graph pass through (0, -3)

So, we have:

-3 = a(0 - 1)^2 - 2

Evaluate the like terms

a = -1

Substitute a = -1 in y = a(x - 1)^2 - 2

y = -(x - 1)^2 - 2

Hence, the equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2

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The diagram ends at -2

The equation would be 2-4 = -2

Answer:

Step-by-step explanation:

Start on point 2

Ends on point -2

Then:

2 + a = -2

a = -2 - 2

a = -4

then the equation is:

2 + (-4) = 0

2 - 4 = 0

As currently written, solving for [tex]x[/tex], we have

[tex]\sqrt{15} - x + \sqrt3 - x = 6[/tex]

[tex]2x = \sqrt{15} + \sqrt3 - 6[/tex]

[tex]x = \dfrac{\sqrt{15} +\sqrt3 - 6}2[/tex]

I suspect you meant to write

[tex]\sqrt{15 - x} + \sqrt{3 - x} = 6[/tex]

Move one of the roots to the other side, then take squares on both sides.

[tex]\sqrt{15 - x} = 6 - \sqrt{3 - x}[/tex]

[tex]\left(\sqrt{15 - x}\right)^2 = \left(6 - \sqrt{3 - x}\right)^2[/tex]

[tex]15 - x = 36 - 12 \sqrt{3 - x} + (3 - x)[/tex]

[tex]12 \sqrt{3 - x} = 24[/tex]

Take squares again

[tex]\left(12\sqrt{3-x}\right)^2 = 24^2[/tex]

[tex]144 (3 - x) = 576[/tex]

[tex]432 - 144x = 576[/tex]

[tex]144x = -144[/tex]

[tex]\boxed{x = -1}[/tex]

Answer:

Perimeter: 4+[tex]\pi \\[/tex] mm

Area: [tex]\pi[/tex] mm^2

Step-by-step explanation:

Perimeter:

90/360 * 4[tex]\pi \\[/tex] +2 + 2 = 4+ [tex]\pi \\[/tex]

Area:

90/360 * 4[tex]\pi[/tex] = [tex]\pi[/tex]

Answer: 7.14 mm and 3.14 mm²

Step-by-step explanation:

The perimeter of a figure is the total length on the outside of the figure. A regular circle has a perimeter of [tex]2\pi r[/tex], where r is the radius and [tex]\pi[/tex] is an irrational constant, approximately 3.14.

Due to the right angle, we know that this is a quarter circle, as it is a quarter of 360°, or a full circle. Hence, the curved portion of the quarter circle is [tex]\frac{1}{4}* 2\pi r[/tex] or [tex]\frac{1}{2} \pi r[/tex]. The radius is 2 mm, so this value is

[tex]\frac{1}{2} \pi*2=\pi[/tex]

However, this isn't the total perimeter, as there are straight edges in the figure too. Both edges are radii of the whole circle, so both would be 2mm. Their total is

[tex]2+2=4[/tex]

By adding [tex]\pi[/tex] to 4 we get [tex]\pi +4[/tex] or 7.14 mm.

The  area is the total space enclosed by a figure. The total area of a whole circle is [tex]\pi r^2[/tex], where r is still the radius. Since this is a quarter circle, it would take up a quarter of the area, or [tex]\frac{1}{4} \pi r^2[/tex]. We can plug in 2 for r and solve to get the area.

[tex]\frac{1}{4}\pi*2^2\\\frac{1}{4}\pi*4\\\pi[/tex]

Hence, the area is [tex]\pi[/tex], or around 3.14 mm².

Question 2

1) [tex]\overline{PR} \perp \overline{QS}[/tex], [tex]\overline{PQ} \cong \overline{PS}[/tex] (given)

2) [tex]\overline{PR} \cong \overline{PR}[/tex] (reflexive property)

3) [tex]\angle PRQ[/tex] and [tex]\angle PRS[/tex] are right angles (perpendicular lines form right angles)

4) [tex]\triangle PRQ[/tex] and [tex]\triangle PRS[/tex] are right triangles (a triangle with a right angle is a right triangle)

5) [tex]\triangle PRQ \cong \triangle PRS[/tex] (HL)

Question 3

1) [tex]\angle C[/tex] is a right angle, [tex]\overline{AC} \cong \overline{AE}[/tex], [tex]\overline{DE} \perp \overline{AB}[/tex] (given)

2) [tex]\angle DEA[/tex] is a right angle (perpendicular lines form right angles)

3) [tex]\triangle ACD[/tex] and [tex]\triangle DAE[/tex] are right triangles (a triangle with a right triangle is a right angle)

4) [tex]\overline{AD} \cong \overline{AD}[/tex] (reflexive property)

5) [tex]\triangle ACD \cong \triangle AED[/tex] (HL)

6) [tex]\angle CAD \cong \angle DAE[/tex] (CPCTC)

7) [tex]\overline{AD}[/tex] bisects [tex]\angle BAC[/tex] (if a segment splits an angle into two congruent parts, it is an angle bisector)

Answer: The answer is factor 3 out the variable terms

Step-by-step explanation:

Answer:

See below

Step-by-step explanation:

Looks like positive linear to me....slope is up and to the right (positive) and the line drawn shows it ot be approx linear

The missing angle of the right trapezoid is 126°.

There are different types of quadrilaterals, for example, square, rectangle, rhombus, trapezoid, and parallelogram.  Each type is defined accordingly to its length of sides and angles. The trapezoid is a quadrilateral that presents at least one set of parallel sides and the sum of the internal angles is 360°. The trapezoid classifications are:

From the figure, it is possible to see that the image has two right angles. Thus, you have a right trapezoid.

Therefore, from the sum of internal angles, you can find the missing angle:

u+54+90+90=360

u=360-234

u=126

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Thus, distribution is a function that shows the possible values for a variable and how often they occur and categorical values are summarized by counts and/or percents

In statistics, the distribution is a function that shows the possible values for a variable and how often they occur.

The distribution of a data set is the shape of the graph when all possible values are plotted on a frequency graph. Usually, we are not able to collect all the data for our variable of interest. Therefore we take a sample. This sample is used to make conclusions about the whole data set.

The categorical variable is a data in which individuals are placed into groups or categories.

One way to summarize categorical data is to simply count (frequency), or tally up, the number of individuals that fall into each category. Another way is to show the percentage of individuals who fall into each category, thereby creating a relative frequency.

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The machine produces 3000 nails in the month of April 2007.

Production of Nails in the Month of April

According to the given information,

Number of nails manufactured per day by the machine, x = 300  

And, we know that April is the fourth month of a year which has total 30 days.

Hence, number of days the machine is manufacturing nails in the month of April, y = 30

So, the number of nails that will be produced by the machine in the month of April is given by,

N = Number of nails produced per day × Number of days in April

N = xy

N = 300 × 30

N = 9000

Therefore, the machine produces 9000 nails in the April month of year 2007.

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#1

h(2.5)

After 2.5s the ball was at 47ft

#2

g(0)

g(4)

g(7)

Range

#3

Answer:

  (b)  I, III

Step-by-step explanation:

The correct answer can be chosen based on your knowledge of the shape of the graph of f(x).

The leading coefficient of this quadratic function being positive tells you the graph will be a parabola that opens upward. The left branch of the parabola will extend to positive infinity, as will the right branch.

If there are x-intercepts, the x-values between those will be where the graph has crossed the x-axis and function values are negative.

The answer choices suggest that x=2 and x=5 are x-intercepts of the function's graph. These can be checked, if you like, by evaluating f(2) and f(5).

  f(2) = 2² -7·2 +10 = 4 -14 +10 = 0

  f(5) = 5² -7·5 +10 = 25 -35 +10 = 0

This means the function will be positive for x < 2 and for x > 5. These intervals are described by I and III.

The value of f from the given expression is 3

Given the expression below

cos(22f-1) = sin(7f+4),

This can also be expressed

cos(22f-1) = cos(90-7f-4)

cos(22f-1) = cos(86-7f)

22f - 1 = 86 - 7f

Collect the like terms

22f+7f = 86 + 1

29f = 87

f = 3

Hence the value of f from the given expression is 3

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Answer:

f-¹(x) = (x + 6)/2

Step-by-step explanation:

Replace f(x) with y.

y = 2x – 6

Interchange x and y.

x = 2y – 6

Solve for y.

y = (x + 6)/2

Replace y with  f-¹(x)

f-¹(x) = (x + 6)/2

Answer:

1

Step-by-step explanation:

[tex]\cos A=\frac{3}{4.24} \\ \\ \cos B=\frac{3}{4.24} \\ \\ \frac{\cos A}{\cos B}=1[/tex]

Answer:

Breadth=8

Length=6

OR

breadth = 6

length=8

Step-by-step explanation:

Area of 4 walls = 2(l+b)h

112=2.4(l+b)

112/8=l+b

14=l+b

l=14-b

Area of base or ceiling=l x b

48=(14-b).b

48=14b-b²

b²-14b+48=0

factorize it and we get

(b-6)(b-8)=0

taking b-6=0

b=6

l=14-6

l=8

The amount of paint needed to cover the wall is 112 square feet

The given parameters are:

Parallel bases = 8 feet and 20 feet

Height = 8 feet

The area is calculated as:

Area = 0.5 * (Sum of parallel bases) * Height

So, we have;

Area = 0.5 * (8 + 20) * 8

Evaluate

Area = 112

Hence, the amount of paint needed to cover the wall is 112 square feet

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The percentage of great lakes states would we expect the economic impact from fishing to be between $234.5 and $401.5 is 50%.

Given that mean is $318 , standard deviation is $83.5 and confidence interval is $234.5 and $401.5.

We have to calculate the percentage of great  lakes that we would expect the economic impact from fishing to be between $234.5 and $401.5.

Because we have not given the sample size taken so we will use any test.We are using z test.

We know that,

z=X-μ/σ

where μ is population mean and σ is population standard deviation.

We cannot find p value between $234.5 and $401.5. We have to find the p value between $234.5 and $318 and then add the p value between $318 and $401.5 by calculating.

P value between $234.5 and $318:

z=234.5-318/83.5

=-83.5/83.5

=-1

p value of -1 is 0.1587

P value between $318 and $401.5:

z=401.5-318.5/83.5

=83.5/83.5

=1

p value of z=1 is 0.3413

Total p value=0.1587+0.3413

=0.5

Total percentage=0.5*100

=50%

Hence the percentage that we would expect the impact to be between $234.5 and $401.5 is 50%.

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The surface area of the composite figure is 524 square meters

The given composite figure is made up of a triangular prism and a rectangular prism. The surface area of the figure is given as:

Surface area = 2(16*5+5*4+16*4) + (16*6) + 2(50)

Take the sum

Surface area = 2(80+20+64) + 96 + 100

Surface area = 2(164) +196

Surface area = 328 + 196

Surface area = 524 square meters

Hence the surface area of the composite figure is 524 square meters

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Step-by-step explanation:

we use the point-slope approach :

y - y1 = m(x - x1)

m is the slope, and (x1, y1) is a point on the line.

so,

1.

y - 2 = 7(x - 1) = 7x - 7

y = 7x - 5

2.

y - -1 = -1×(x - 3) = -x + 3

y + 1 = -x + 3

y = -x + 2

3.

y - 5 = -4(x - -2) = -4x - 8 (3x a "-" is a "-" for 8)

y = -4x - 3

4.

y - 5 = 5/3 × (x - 3) = 5x/3 - 5

y = 5x/3 or 5/3 × x

Answer:

1. [tex]7x-y=5[/tex]

2. [tex]x+y=2[/tex]

3. [tex]4x+y=-3[/tex]

4. [tex]5x-3y=0[/tex]

Step-by-step explanation:

Since we already know the slope and the coordinates of one point, first use the point-slope form to create an equation.

[tex]y-k=m(x-h)[/tex]

[tex]y-2=7(x-1)[/tex]

[tex]y-2=7x-7[/tex]

[tex]7x-y=5[/tex]

[tex]y+1=-1(x-3)[/tex]

[tex]y+1=-x+3[/tex]

[tex]x+y=2[/tex]

[tex]y-5=-4(x+2)[/tex]

[tex]y-5=-4x-8[/tex]

[tex]4x+y=-3[/tex]

[tex]y-5=\frac{5}{3} (x-3)[/tex]

[tex]y-5=\frac{5}{3}x-5[/tex]

[tex]3y-15=5x-15[/tex]

[tex]5x-3y=0[/tex]

[tex] \frac{y}{x} = \frac{1}{ - 3} \\ - 3y = x[/tex]

[tex] - 3(2) = 6 = x[/tex]