Show, analytically, that the equation sec∝ + tan ∝ / sec ∝ - tan ∝ = 1+ 2sin∝ + sin²∝ /cos²∝ is an identity for all values of ∝ on the domain of each expression.

Answer:

d. All of the above are true

Step-by-step explanation:

All of the following statements,

a. the stronger the linear relationship between two variables, the closer the correlation coefficient will be to 1.

b. if the correlation coefficient between the x and y variables is negative, the sign on the regression slope will also be negative.

C. if the correlation coefficient between the dependent and independent variable is determined to be significant, the regression model for y given x will also be significant.

are true

The problem involves the calculation of the total distance covered by the car.A car that moves from rest acquires a velocity of 20 m/s with uniform acceleration in 4 seconds. This means that its acceleration was 5 m/s², and it moved 20 m/s after 4 seconds.

From this moment, the car moved with this velocity for 6 seconds. This means that it covered a distance of 20 m/s × 6 s = 120 m. After 10 seconds, the car has covered a distance of 120 m. It then accelerates uniformly to 30 m/s in 5 seconds. The acceleration of the car is (30 – 20) m/s ÷ 5 s = 2 m/s².

Therefore, the car would have covered a distance of 20 m/s × 5 s + 0.5 × 2 m/s² × (5 s)² = 75 m during the acceleration phase of 5 seconds. When it travels for 3 seconds at this velocity, it would have covered a distance of 30 m/s × 3 s = 90 m.

Adding the distance covered during the first 10 seconds to the distance covered during the next 8 seconds, we have a total distance of 120 m + 75 m + 90 m = 285 m.When the car comes to rest, it does so with uniform declaration in 12 seconds. The acceleration of the car at this time is -2.5 m/s².

Therefore, the car would have covered a distance of 30 m/s × 12 s + 0.5 × (-2.5 m/s²) × (12 s)² = 180 m when coming to a stop.Thus, the total distance covered by the car is 285 m + 180 m = 465 m. Therefore, the total distance covered by the car is 465 meters.

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The Geometric Mean of a Leg Theorem, or the Geometric Mean Theorem, is related to right triangles and their altitude.

The Geometric Mean of a Leg Theorem states that " In a right triangle, the length of the altitude to the hypotenuse is the geometric mean of the lengths of the two segments of the hypotenuse created by the altitude."

It can be proven by assuming you have a right  triangle ABC, where angle BAC is the right angle, BC is the hypotenuse, AD is the altitude, and BD and DC are the two segments of the hypotenuse created by the altitude.

This shows the proof of the Geometric Mean of a Leg Theorem.

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Answer:

2a2b-6ab2+9ab+5 is right answer

Answer:

\( \frac{1}{2} 3 \\ \)

Step-by-step explanation:

Your welcome

The answer is D for the first and B for the second.

Answer:

5/8m+3/8

Step-by-step explanation:

1/4m+3/4m-3/8(m+1)

1/4m+3/4m-3/8m+3/8

5/8m+3/8

Answer:

5/8m + 3/8

Step-by-step explanation:

1/4m + 3/4m - 3/8 (m + 1)

1/4m + 3/4m - 3/8m + 3/8

5/8m + 3/8

Answer:

the answer is 3

Step-by-step explanation:

3=4

Based on the provided information, the mean of the area of the circle is 1/3 and the variance is 4/45.

The mean and variance of the area of a circle with a uniformly distributed radius on the interval (0,1) can be found using the expected value and variance formulas for continuous random variables.

The expected value (mean) of a continuous random variable X is given by:

E[X] = ∫xf(x)dx

Where f(x) is the probability density function of X. In this case, since the radius is uniformly distributed on the interval (0,1), the probability density function is f(x) = 1 for 0 ≤ x ≤ 1.

The expected value of the area of the circle is therefore:

E[A] = ∫a*f(a)da = ∫r^2*1dr = (1/3)r^3 for 0 ≤ r ≤ 1 = (1/3)(1)^3 - (1/3)(0)^3 = 1/3

The variance of a continuous random variable X is given by:

Var[X] = E[X^2] - (E[X])^2

The expected value of the square of the area of the circle is:

E[A^2] = ∫a^2*f(a)da = ∫r^4*1dr = (1/5)r^5 for 0 ≤ r ≤ 1 = (1/5)(1)^5 - (1/5)(0)^5 = 1/5

Therefore, the variance of the area of the circle is:

Var[A] = E[A^2] - (E[A])^2 = 1/5 - (1/3)^2 = 4/45

So the mean of the area of the circle is 1/3 and the variance is 4/45.

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Let the integer be X

2x+64=99

2x= 99-64

2x= 34

x=34÷2

X= 17.5

Answer:

x=30

Step-by-step explanation:

x+90+60=180

x=180-150

x=30

Answer:

d-30

Step-by-step explanation:

Answer:

y = 24/x

Step-by-step explanation:

\(\qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}} ~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\cfrac{\stackrel{\downarrow }{k}}{x}~\hfill } \\\\ \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] ~\dotfill\)

\(\stackrel{\textit{"y" varies inversely with "x"}}{y = \cfrac{k}{x}}\qquad \textit{we also know that} \begin{cases} y = 4\\ x = 6 \end{cases} \\\\\\ 4=\cfrac{k}{6}\implies 24=k~\hspace{10em}\boxed{y = \cfrac{24}{x}}\)

Given:

Loan = P = $48,000

Simple interest rate = r = 2% = 0.02

Time = t = 9 years

So,

The interest = I =

Total amount =

To find the monthly payments, we will divide the total amount over the number of months of the nine years

so,

The monthly payments =

We calculated the probabilities of specific events such as the sum being less than or equal to 9, the sum being 7, and the sum being an even number.

To construct the probability distribution table, we need to calculate the probabilities of each possible outcome when two fair dice are tossed simultaneously. The sum of the two numbers can range from 2 to 12.

(a) Probability distribution table:

Sum (x) Probability (P(x))

2 1/36

3 2/36

4 3/36

5 4/36

6 5/36

7 6/36

8 5/36

9 4/36

10 3/36

11 2/36

12 1/36

(b) Probability of getting a sum less than or equal to 9:

To calculate this probability, we sum the probabilities of getting each sum from 2 to 9:

P(x ≤ 9) = P(x = 2) + P(x = 3) + ... + P(x = 9) = 1/36 + 2/36 + ... + 4/36

= 20/36 = 5/9

(c) Probability of getting a sum of 7:

P(x = 7) = 6/36 = 1/6

(d) Probability of getting an even number:

To calculate this probability, we sum the probabilities of getting sums 2, 4, 6, 8, 10, and 12:

P(even) = P(x = 2) + P(x = 4) + ... + P(x = 12) = 1/36 + 3/36 + ... + 1/36

= 18/36 = 1/2

(e) Mean:

To calculate the mean, we multiply each sum by its corresponding probability and sum the results:

Mean = (2 * 1/36) + (3 * 2/36) + ... + (12 * 1/36)

= 7

(f) Standard Deviation:

The standard deviation can be calculated using the formula:

Standard Deviation = √[ Σ(x - μ)^2 * P(x) ]

where μ is the mean.

(g) Variance:

Variance = Standard Deviation^2

Calculating the standard deviation and variance requires additional calculations that cannot be done within the text-based interface. However, using the probabilities and mean provided, the calculations can be done using appropriate software or a calculator.

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No, when finding the upper and lower quartiles, the median is not included in the calculations.

When finding the upper and lower quartiles of a data set, the median (or second quartile) is not included in the calculations. The quartiles divide the data set into four equal parts, with the median representing the second quartile.

To find the lower quartile (Q1), one needs to determine the median of the lower half of the data set, excluding the median itself. This includes the data points below the median.

Similarly, to find the upper quartile (Q3), the median of the upper half of the data set is determined, excluding the median. This includes the data points above the median.

The inclusion of the median in the calculation of quartiles can cause confusion, but it is important to note that the median is separate and distinct from the quartiles.

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Answer:

y = 2/3x + 5

Step-by-step explanation:

y = mx + b is slope intercept form

You have the equation in order but in order to complete it, you need to remove the 3 in front of the y.

#1: 3y = 2x + 15 (Remove the 3 by diving all the numbers on the opposite side)

Final Answer: y = 2/3x + 5

Answer: y=2/3x-5

Step-by-step explanation:

Rearrange 2x−3y=15 into this form

subtract 2x from both sides.

divide ALL terms on both sides by - 3

⇒ y=2/3x-5

Answer:

a.option is the correct answer

Answer:

I did it and is 36

Step-by-step explanation:

Answer:

The first one: 28

Step-by-step explanation:

The total number of people who like cooking is 20 people while 48 people like to read.

48-20 is 28

Scientific notation is the way through which a very small or a very large number can be written in shorthand. The given following number or sentences can be written in the form of scientific notation as shown.

Scientific notation is the way through which a very small or a very large number can be written in shorthand.  In scientific notations when a number between 1 and 10s is multiplied by a power of 10.

For example, 6,500,000,000,000 can be written as 6.5e+12 or 6.5×10¹².

The given following number or sentences can be written in the form of scientific notation as shown below.

a.) 1,000,000 = 1×10⁶ = 1e+6

b.) .00005 = 5×10⁻⁵ = 5e-5

c.) 40 thousand = 40,000 = 4×10⁴ = 4e+4

d.) 7 trillion = 7,000,000,000,000 = 7×10¹² = 7e+12

e.) 8 thousandths = 0.008 = 8×10⁻³ = 8e-3

f.) one tenth = 0.1 = 1×10⁻¹ = 1e-1

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Answer:

Step-by-step explanation:

Hope this helps :)

With the Hubble's constant H₀, the estimated age of the universe would be:

T = 8,332,452,617.49 years.

We know that the age of the universe is something near to the time the galaxies needed to reach their current distance:

T = D/V

And by Hubble's law, we know that:

V = H₀*D

Then we can write:

T = D/(H₀*D) = 1/H₀

So, we can say that the age of the universe is something near the inverse of Hubble's constant.

Then we have:

T = 1/(36 km/s*Mly) = (1/36)  s*Mly/km

Now we need to perform the correspondent change of units.

1 Mly = 1 million light-years

Such that:

1 ly = 9.461*10^12 km

Then 1 million light-years over km is equal to:

1 Mly/km = 1,000,000*(9.461*10^12 km)/km = 9.461*10^18

Then we can replace it:

T = (1/36) s*Mly/km = (1/36)*9.461*10^18 s

T = 2.628*10^17 s

This is the age in seconds, but we want it in years.

We know that:

1 year = 3.154*10^7 s

Then to change the units, we compute:

T = (2.628*10^17 s/3.154*10^7 s)* 1 yea

T = 8,332,452,617.49 years.

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Quadrant Class 9 is a level of mathematics taught in secondary school in the United Kingdom. It is the equivalent of Algebra 1 and Geometry combined in the United States.

This class teaches students how to graph equations, solve equations, and use the properties of shapes to solve problems. Quadrant Class 9 also covers the basics of calculus, such as derivatives, integrals, and limits.

The main focus of Quadrant Class 9 is on graphing equations. Students learn how to plot points on a graph, connect the points to make a line or curve, and solve equations by finding the intersection of two lines. They also learn how to graph the solutions of inequalities and identify the domain and range of a function.

In addition to graphing, students learn how to solve equations using basic algebraic principles, such as the order of operations and the distributive property. They also learn how to solve systems of equations, manipulate polynomials, and find the roots of a quadratic equation.

Quadrant Class 9 also covers the basics of geometry, such as the Pythagorean Theorem, the properties of triangles, and the area and perimeter of shapes. Students learn how to use these skills to solve problems involving angles, area, and volume.

Overall, Quadrant Class 9 is a challenging and rewarding mathematics class that prepares students for higher-level math classes in the future.

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Answer:

4.25%

Step-by-step explanation:

Commision =  441.30 - 375 = 66.30

Commision is based on  = 6560-5000 = 1560

Rate of Commision = (in decimal) 66.30/1560 = 0.0425  

Rate of Commision = 0.0425  * 100 =   4.25 %

Answer:

\( \theta = 51.8^\circ \)   or   \( \theta = 308.2^\circ \)

Step-by-step explanation:

\( \cos^2 \theta + \cos \theta − 1 = 0 \)

\( x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \cos \theta = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} \)

\( \cos \theta = \dfrac{-1 \pm \sqrt{1 + 4}}{2} \)

\( \cos \theta = \dfrac{-1 \pm \sqrt{5}}{2} \)

\( \cos \theta = 0.61803 \)   or  \( \cos \theta = -1.61803 \)

The range of the cos θ function excludes θ = -1.61803, so we discard that solution.

\( \theta = 51.8^\circ \)   or   \( \theta = 308.2^\circ \)

Answer:

4 inches of snow has fallen rate per hour

therefore the unit rate is 4 inches of snow per hr.

Step-by-step explanation:

Answer:

4 per hour

Step-by-step explanation:

because 12 divided by 3 is 4

Answer:

Step-by-step explanation:

Answer:

WYM

Step-by-step explanation:

I think the guy who see's the best is on dah chair

f(n) = ϴ(g(n)) is not true in cases B(sqrt(n)log(n), C(\(3^n 5^n\)), and D(sin(n)+3 cos(n)+1).

A) \(log(n^200) log(n^2)\)

Here, f(n) = \(log(n^200)\) and g(n) = \(log(n^2)\). Now, if we take the limit of f(n) / g(n) as n approaches infinity, then:

f(n) / g(n) = \([log(n^200) / log(n^2)]\) = 100

This means that as n approaches infinity, the ratio f(n) / g(n) is constant, and so we can say that f(n) = ϴ(g(n)). Therefore, f(n) = ϴ(g(n)) is true in this case.

B) sqrt(n) log(n) Here, f(n) = sqrt(n) and g(n) = log(n). Now, if we take the limit of f(n) / g(n) as n approaches infinity, then:

f(n) / g(n) = [sqrt(n) / log(n)]

As log(n) grows much slower than sqrt(n) as n approaches infinity, this limit approaches infinity. Therefore, we cannot say that f(n) = ϴ(g(n)) is true in this case.

C) 3^n 5^n

Here, f(n) = \(3^n\) and g(n) = \(5^n\) . Now, if we take the limit of f(n) / g(n) as n approaches infinity, then:

f(n) / g(n) = \([3^n / 5^n]\)

As \(3^n\) grows much slower than \(5^n\) as n approaches infinity, this limit approaches zero. Therefore, we cannot say that f(n) = ϴ(g(n)) is true in this case.

D) sin(n) + 3 cos(n) + 1

Here, f(n) = sin(n) + 3 and g(n) = cos(n) + 1. Now, if we take the limit of f(n) / g(n) as n approaches infinity, then:

f(n) / g(n) = [sin(n) + 3] / [cos(n) + 1]

As this limit oscillates between positive and negative infinity as n approaches infinity, we cannot say that f(n) = ϴ(g(n)) is true in this case.

Therefore, f(n) = ϴ(g(n)) is not true in cases B, C, and D.

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Step-by-step explanation:

To find the circumference of a circle you use 2 · π · r