Write the inequality represented on each graph. Use n for the variable.

The speed of light is 300,000 kilometers per hour. hence option c is correct.

According to the statement

we have to explain that the by how much speed the light travels.

So, Firstly

Light is the natural agent that stimulates sight and makes things visible.

and speed of light means that the distance covered by the light with respect to time.

So, it means those distance which is covered by the light with respect to time is called the speed of light.

The speed of light is measures in km, meters and in other units also.

And we can calculate the speed of light per hour or per minute or in other time measurements.

So, The speed of light is 300,000 kilometers per hour. hence option c is correct.

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1. The time taken for the water to fall that much is 0.2 s

2. The initial speed of the water is 4 m/s

h = ½gt²

0.2 = ½ × 9.8 × t²

0.2 = 4.9 × t²

Divide both side by 4.9

t² = 0.2 / 4.9

Take the square root of both side

t = √(0.2 / 4.9)

t = 0.2 s

s = ut

Divide both sides by t

u = s / t

u = 0.8 / 0.2

u = 4 m/s

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The given function for the difference in length is presented as follows;

[tex]f( \theta) = 2 \cdot cos( \theta) + \sqrt{3} [/tex]

When the pogo stick will be equal to its non compressed length, the difference is zero, therefore;

[tex]f( \theta) = 2 \cdot cos( \theta) + \sqrt{3} = 0[/tex]

[tex] 2 \cdot cos( \theta) = - \sqrt{3} [/tex]

[tex]\theta= arccos \left( \frac{ - \sqrt{3} }{2} \right) [/tex]

Which gives;

[tex] \theta = \frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{6} [/tex]

[tex] \theta = -\frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{6} [/tex]

Part B; If the angle was doubled, we have;

[tex]f( \theta) = 2 \cdot cos(2 \cdot \theta) + \sqrt{3} = 0[/tex]

Therefore;

[tex] 2 \cdot cos(2 \cdot \theta) = - \sqrt{3} [/tex]

Which gives;

[tex] \theta = \frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{12} [/tex]

[tex] \theta = -\frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{12} [/tex]

Between 0 and 2•π, we have;

[tex] \theta = \frac{5\cdot \pi}{12} [/tex]

[tex] \theta = \frac{ 17\cdot \pi}{12} [/tex]

Part C;

The toddler's pogo stick is presented as follows;

[tex]g( \theta) = 1- son^2( \theta) + \sqrt{3} [/tex]

Integrating the original function between 0 and theta gives;

[tex]2 \cdot sin( \theta) + \sqrt{3}\cdot \theta [/tex]

The original length =

Therefore, when the lengths are equal, we have;

[tex]1- son^2( \theta) + \sqrt{3} = 2 \cdot sin( \theta) + \sqrt{3}\cdot \theta [/tex]

The two equivalent forms of [tex]sec^2x[/tex] are [tex]\frac{1}{cos^2x}[/tex] and [tex]1+tan^2x[/tex]

The given trigonometric expression is:

[tex]sec^2x[/tex]

Note that:

[tex]sec(x)=\frac{1}{cos(x)}[/tex]

Substitute this equivalence to the given expression

[tex]sec^2(x)=\frac{1}{cos^2(x)}[/tex]

Also from [tex]cos^2(x)+sin^2(x)=1[/tex]

Divide through by [tex]cos^2x[/tex]

[tex]\frac{cos^2x}{cos^2x} +\frac{sin^2x}{cos^2x}=\frac{1}{sin^2x}[/tex]

Simplifying the resulting expression:

[tex]1+tan^2x=sec^2x\\\\sec^2x=1+tan^2x[/tex]

Therefore, the two equivalent forms of [tex]sec^2x[/tex] are [tex]\frac{1}{cos^2x}[/tex] and [tex]1+tan^2x[/tex]

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The test that represents the best choice if you wanted to compare the average mean time to repair (mttr) of your office for the last month to the division’s mttr is one sample t-test

The purpose of the one sample t-test is to determine if a sample observations could follow a specific parameter like the mean.

This test is used to determine whether an unknown population mean is different from a specific value.

one sample t-test is used to compare the mean of a single group against that of a known mean.

We wanted to compare the average mean time to repair (mttr) of your office for the last month to the division’s mttr

So this test (one sample t-test ) would be the best choice to compare mttr.

Therefore, the test that represents the best choice if you wanted to compare the average mean time to repair (mttr) of your office for the last month to the division’s mttr is one sample t-test

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a

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The average increase in the number of flowers pollinated per day between days 4 and 10 is 39, given that the number of pollinated flowers as a function of time in days can be represented by the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex].

In the question, we are asked for the average increase in the number of flowers pollinated per day between days 4 and 10, given that the number of pollinated flowers as a function of time in days can be represented by the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex].

To find the average increase in the number of flowers pollinated per day between days 4 and 10, we use the formula {f(10) - f(4)}/{10 - 4}.

First, we find the value of the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex], for f(10) and f(4).

[tex]f(x) = (3)^{\frac{x}{2} }\\\Rightarrow f(10) = (3)^{\frac{10}{2} }\\\Rightarrow f(10) = 3^5 = 243[/tex]

[tex]f(x) = (3)^{\frac{x}{2} }\\\Rightarrow f(4) = (3)^{\frac{4}{2} }\\\Rightarrow f(10) = 3^2 = 9[/tex]

Thus, the average increase

= {f(10) - f(4)}/{10 - 4},

= (243 - 9)/(10 - 4),

= 234/6

= 39.

Thus, the average increase in the number of flowers pollinated per day between days 4 and 10 is 39, given that the number of pollinated flowers as a function of time in days can be represented by the function [tex]f(x) = (3)^{\frac{x}{2} }[/tex].

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For complete question, refer to the attachment.

Answer:

Step-by-step explanation:

From the right triangle with cot theta = 4/7  = (adjacent side / opposite side), Fatima can find the length of the hypotenuse using Pythagoras. This gives  a value of 8.0823. So cos theta = 4 / 8.0823 = 0,49614,

Then she could find the values of sin theta  using

sine squared theta + cos squared theta = 1.

Answer: its D or 1 + cotangent squared theta = cosecant squared theta

Step-by-step explanation:

The Polynomial Roots are given as follows:
 

x₁=−1.2085

x₂=0.2709+1.2576i

x₃=0.2709−1.2576i

It is to be noted that the Polynomial 3x³+ 2x2 + 3x + 6 has no real rational roots that can be found using the rational root text. Hence, this was solved using the Qubic Formulas.

The roots (also known as zeroes or solutions) of a polynomial P (x) P(x) P(x) are the x values for which P (x) P(x) P(x) equals zero.

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​

Answer: 12

Step-by-step explanation:

I got it right

The answers to all the parts are given below.

The height equation is:

h(t) = - 16t^2 + 100t + 10

A) To find the maximum height, we must find the time where the velocity is 0, so we must derive it with respect to the time.

v(t) = - 2 * 16 * t + 100 = 0

t = 100/(2*16) = 3.125s

Now we put this time in our height equation:

h(3.125s) = -16*3.125^2 + 100*3.125 + 10 = 166.25 ft.

B) The lowest height of the potato will be h = 0ft when the potato hits the ground.

C) the lowest time that works for that function is t = 0s when the potato is fired by the gun.

D) the maximum time can be found when the potato hits the ground, after that point the equation does not work anymore, let's find it,

h(t) = 0 = -16t^2 +100t+10

we can solve it using Bhaskara's equation:

[tex]t=\frac{-100+-\sqrt{100^{2} +4*16*10} }{-2*16} =\frac{-100+-103.2}{-32}[/tex]

So the two solutions are:

t = 6.3s

t = -0.1s

We need to choose the positive time, as we already discussed that the minimum time that works for the equation is t = 0s.

so here the answer is t = 6.3s

E) the domain is  0s ≤ t ≤ 6.3s

The range is 0ft ≤ h ≤ 166.25 ft.

Therefore, the answers to all the parts are shown.

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Answer:

1st option

Step-by-step explanation:

let y = h(x) and rearrange making x the subject , that is

y = 6x - 6 ( add 6 to both sides )

y + 6 = 6x ( divide both sides by 6 )

[tex]\frac{y+6}{6}[/tex] = x

change y back into terms of x with x = [tex]h^{-1}[/tex] (x)

[tex]h^{-1}[/tex] (x) = [tex]\frac{x+6}{6}[/tex]

Question 4

[tex]\sin 30^{\circ}=\frac{y}{14\sqrt{3}}\\\\\frac{1}{2}=\frac{y}{14\sqrt{3}}\\\\\boxed{y=7\sqrt{3}}\\\\\\\\\cos 30^{\circ}=\frac{x}{14\sqrt{3}}\\\\\frac{\sqrt{3}}{2}=\frac{x}{14\sqrt{3}}\\\\\boxed{x=21}[/tex]

Question 5

[tex]\cos 45^{\circ}=\frac{x}{28}\\\\\frac{1}{\sqrt{2}}=\frac{x}{28}\\\\x=\frac{28}{\sqrt{2}}\\\\\boxed{x=14\sqrt{2}}\\\\\\\\\sin 45^{\circ}=\frac{y}{28}\\\\\frac{1}{\sqrt{2}}=\frac{y}{28}\\\\y=\frac{28}{\sqrt{2}}\\\\\boxed{y=14\sqrt{2}}[/tex]

Answer:

area of A'B'C'D' = 18.4 units²

Step-by-step explanation:

given 2 similar figures with

ratio of corresponding sides = a : b , then

ratio of areas = a² : b²

here ratio of sides = 5 : 2

ratio of areas = 5² : 2² = 25 : 4

let x be the area of A'B'C'D' , then by proportion

[tex]\frac{25}{115}[/tex] = [tex]\frac{4}{x}[/tex] ( cross- multiply )

25x = 460 ( divide both sides by 25 )

x = 18.4

area of A'B'C'D' = 18.4 units²

The high school graduation is the dependent variable.

In this question,

Jordan is studying whether or not race affects high school graduation in utah.

A dependent variable is a variable in an expression that depends on the value of another variable. It is a variable that represents a quantity that changes based on other quantities being manipulated in an experiment. It is the variable being tested, and therefore, it is called the dependent variable.

In the above statement, the high school graduation gets affected by race. So, the the high school graduation is the dependent variable.

Hence we can conclude that the high school graduation is the dependent variable.

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Answer:

35449 people

Explanation:

Given equation: y = 1100.74x -1976.47

Here year '2005' represents x = 005 = 5

So, the year '2034' will represent x = 034 = 34

Insert this into equation:

y = 1100.74(34) -1976.47

y = 35448.69

y ≈ 35449 (rounded to nearest whole number)

Answer:

[tex]\sf A) \quad \begin{cases}\sf y > -5x+5\\\sf y < -5x+12\end{cases}[/tex]

B)  see below

C)  points C, E and F

Step-by-step explanation:

Given points:

A system of inequalities is a set of two or more inequalities in one or more variables.

To create a system of inequalities that only contains C and F in the overlapping shaded region, create a linear equation where points C, F and E are to the right of the line and a linear equation where points C, F, A, B and D are to the left of the line.

The easiest way to do this is to find the slope of the line that passes through points C and F, then add values to move the lines either side of the points.

[tex]\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{y_F-y_C}{x_F-x_C}=\dfrac{-4-1}{3-2}=-5[/tex]

Therefore:

[tex]\sf y = -5x + 5[/tex]  →  points C, F and E are to the right of the line.

[tex]\sf y=-5x+12[/tex]  →  points C, F, A, B and D are the left of the line.

Therefore, the system of inequalities that only contains points C and F in the overlapping shaded regions is:

[tex]\begin{cases}\sf y > -5x+5\\\sf y < -5x+12\end{cases}[/tex]

To graph the system of inequalities:

To verify that the points C and F are solutions to the system of inequalities created in Part A, substitute the x-values of both points into the system of inequalities.  If the y-values satisfy both inequalities, then the points are solutions to the system.

Point C (2, 1)

[tex]\implies \sf x=2 \implies 1 > -5(2)+5 \implies 1 > -5\quad verified[/tex]

[tex]\implies \sf x=2 \implies 1 < -5(2)+12\implies 1 < 2 \quad verified[/tex]

Point F (3, -4)

[tex]\implies \sf x=3 \implies -4 > -5(3)+5 \implies -4 > -10\quad verified[/tex]

[tex]\implies \sf x=3 \implies -4 < -5(3)+12\implies -4 < -3 \quad verified[/tex]

Method 1

Graph the line y = 7x - 4 (making the line dashed since it is y < 7x - 4).

Shade below the dashed line.

Points that are contained in the shaded region are the houses in which Erica is interested in living:  points C, E and F.

Method 2

Substitute the x-value of each point into the given inequality y < 7x - 4.

Any point where the y-value satisfies the inequality is a house that Erica is interested in living.

[tex]\sf Point\: A: \quad x=-5 \implies 5 < 7(-5)-4 \implies -5 < -39 \implies no[/tex]

[tex]\sf Point\: B: \quad x=-4 \implies -2 < 7(-4)-4 \implies -2 < -32 \implies no[/tex]

[tex]\sf Point\: C: \quad x=2 \implies 1 < 7(2)-4 \implies 1 < 10 \implies yes[/tex]

[tex]\sf Point\: D: \quad x=-2 \implies 4 < 7(-2)-4 \implies 4 < -18 \implies no[/tex]

[tex]\sf Point\: E: \quad x=2 \implies 4 < 7(2)-4 \implies 4 < 10 \implies yes[/tex]

[tex]\sf Point\: F: \quad x=3 \implies -4 < 7(3)-4 \implies -4 < 17 \implies yes[/tex]

A irrational number is a number that can't be expressed as a ratio of two whole numbers. That's it.

For examples (in increasing order of difficulty)

1 is a rational number because it is 1/1

0.75 is a rational number because it is equal to 3/4

2.333... (infinite number of digits, all equal to three) is rational because it is equal to 7/3.

sqrt(2) is not a rational number. This is not completely trivial to show but there are some relatively simple proofs of this fact. It's been known since the greek.

pi is irrational. This is much more complicated and is a result from 19th century.

As you see, there is absolutely no mention of the digits in the definition or in the proofs I presented.

Now the result that you probably hear about and wanted to remember (slightly incorrectly) is that a number is rational if and only if its decimal expansion is eventually periodic. What does it mean ?

Take, 5/700 and write it in decimal expansion. It is 0.0057142857142857.. As you can see the pattern "571428" is repeating in the the digits. That's what it means to have an eventually periodic decimal expansion. The length of the pattern can be anything, but as long as there is a repeating pattern, the number is rational and vice versa.

As a consequence, sqrt(2) does not have a periodic decimal expansion. So it has an infinite number of digits but moreover, the digits do not form any easy repeating pattern.

a.

b.

A quadratic function is a function of the form y = ax² + bx + c

To write y = 5(x - 2)² + 7, we expand the bracket and simplify.

So, y = 5(x - 2)² + 7

y = 5(x² - 4x + 4) + 7

y = 5x² - 20x + 20 + 7

y = 5x² - 20x + 27

So,  the quadratic function y = 5(x - 2)² + 7 in standard form is y = 5x² - 20x + 27

The y intercept is gotten when x = 0.

So, y = 5x² - 20x + 27

y(0) = 5(0)² - 20(0) + 27

y(0) = 0 - 0 + 27

y(0) = 27

The y intercept is (0, 27)

Find the graph in the attachment

To write y = 3(x + 1)² + 12, we expand the bracket and simplify.

So, y = 3(x + 1)² + 12

y = 3(x² + 2x + 1) + 12

y = 3x² + 6x + 3 + 12

y = 3x² + 6x + 15

So, the quadratic function y = 3(x + 1)² + 12 in standard form is y = 3x² + 6x + 15

The y intercept is gotten when x = 0.

So, y = 3x² + 6x + 15

y(0) = 3(0)² - 6(0) + 15

y(0) = 0 - 0 + 15

y(0) = 15

The y-intercept is (0, 15)

Find the graph in the attachment

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Answer:

Step-by-step explanation:

hello :

|x – 5| + 2 < 20  means :  |x – 5| + 2-2 < 20-2

|x – 5|  < 18

-18< x - 5 < 18

-18+5< x - 5+5 < 18+5

-13< x< 23   ( solutions)

The cross-section would be the same shape as the base. A cross-section cannot be a sphere because a cross-section must be two-dimensional, but a sphere is three-dimensional.

This follows from the identity

[tex]\cot(x) = \dfrac1{\tan(x)} = \tan(90^\circ - x)[/tex]

The overall product is 1 because

[tex]\tan(10^\circ) \tan(80^\circ) = \cot(80^\circ) \tan(80^\circ) = 1[/tex]

[tex]\tan(20^\circ) \tan(70^\circ) = \cot(70^\circ) \tan(70^\circ) = 1[/tex]

[tex]\tan(30^\circ) \tan(60^\circ) = \cot(60^\circ) \tan(60^\circ) = 1[/tex]

[tex]\tan(40^\circ) \tan(50^\circ) = \cot(50^\circ) \tan(50^\circ) = 1[/tex]

Answer:

$270

Step-by-step explanation:

18x15=$270

HOPE THAT HELPED:)

The linear inequality of the graph is: -x + 2y + 1 > 0

First, we calculate the slope of the dashed line using:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

Two points on the graph are:

(1, 0) and (3, 1)

The slope (m) is:

[tex]m = \frac{1 - 0}{3 - 1}[/tex]

This gives

m = 0.5

The equation of the line is calculated as:

[tex]y = m(x -x_1) + y_1[/tex]

So, we have;

[tex]y = 0.5(x -1) + 0[/tex]

This gives

[tex]y = 0.5x -0.5[/tex]

Multiply through by 2

[tex]2y = x - 1[/tex]

Now, we convert the equation to an inequality.

The line on the graph is a dashed line. This means that the inequality is either > or <.

Also, the upper region of the graph that is shaded means that the inequality  is >.

So, the equation becomes

2y > x - 1

Rewrite as:

-x + 2y + 1 > 0

So, the linear inequality is: -x + 2y + 1 > 0

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Complete question

Find a linear inequality with the following solution set. Each grid line represents one unit. (Give your answer in the form ax+by+c>0 or ax+by+c [tex]\geq[/tex] 0 where a, b, and c are integers with no common factor greater than 1.)

Answer: 3/4

Step-by-step explanation:

Tangent is opp/adj

Therefore tan (x) = 18/24, which reduces to 3/4

The average rate of change is -5 per month

The interval is given as:

July to December

From the graph, we have:

July = 110

December = 30

The average rate of a function over the interval (a, b) is calculated as:

Rate = [f(b) - f(a)]/[b - a]

So, we have:

Rate = (30 - 110)/(December- July)

This gives

Rate = (30 - 110)/(12 - 7)

Evaluate

Rate = -16

Hence, the average rate of change is -5 per month

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It is an example of Permutation without repetition.

To find the number of pictures can create.

An arrangement of things in a certain sequence is what we refer to as a permutation. The constituents or components of sets are presented in this section in a sequential or chronological fashion. A coming together or merging of several parts or characteristics in which each of the constituent parts or traits retains its unique identity.

Given that:

50 unique pieces of puzzle

Use them all to solve puzzle

No repetition is allowed therefore

Since no repetition is allowed because all puzzle pieces are required to be used.

Therefore each of the 50 pieces can be used max one time

Therefore if we start with piece like this " 1 2 3 .. up to ... 50 " it is one combination

Another could be "2 1 3 ..  up to .. 50" or "3 1 2 .. up to 50" and so on

So we see that we can see that we can select the first piece in 50 ways 49 left which could be selected for second piece then 48, 47 respectively

Therefore 50X49X48X47X46X....X1 = 50!

Hence, It is an example of Permutation without repetition.

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Option is missing

A. Combination with repetition

B. Permutation with repetition

C. Combination without repetition

D. Permutation without repetition