Mathematics High School

xcosa + ysina =p and x sina -ycosa =q​

Answers

Answer 1

We have the value of 'y' in terms of 'x', 'p', 'q', and the trigonometric functions 'sina' and 'cosa'.

To solve the system of equations:

xcosa + ysina = p

xsina - ycosa = q

We can use the method of elimination to eliminate one of the variables.

To eliminate the variable 'sina', we can multiply equation 1 by xsina and equation 2 by xcosa:

x²sina*cosa + xysina² = psina

x²sina*cosa - ycosa² = qcosa

Now, we can subtract equation 2 from equation 1 to eliminate 'sina':

(x²sinacosa + xysina²) - (x²sinacosa - ycosa²) = psina - qcosa

Simplifying, we get:

2xysina² + ycosa² = psina - qcosa

Now, we can solve this equation for 'y':

ycosa² = psina - qcosa - 2xysina²

Dividing both sides by 'cosa²':

y = (psina - qcosa - 2xysina²) / cosa²

So, using 'x', 'p', 'q', and the trigonometric functions'sina' and 'cosa', we can determine the value of 'y'.

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Related Questions

Bearing used in an automotive application is supposed to have a nominal inside diameter 1.5 inches. A random sample of 25 bearings is selected, and the average inside diameter of these bearings is 1.4975 inches. Bearing diameter is known to be normally distributed with standard deviation σ=0.1 inch. We want to test the following hypothesis at α=0.01. H0​:μ=1.5,H1​:μ=1.5 (a) Calculate the type II error if the true mean diameter is 1.55 inches. (b) What sample size would be required to detect a true mean diameter as low as 1.55 inches if you wanted the power of the test to be at least 0.9 ?

Answers

(a) Without knowing the effect size, it is not possible to calculate the type II error for the given hypothesis test. (b) To detect a true mean diameter of 1.55 inches with a power of at least 0.9, approximately 65 bearings would be needed.

(a) If the true mean diameter is 1.55 inches, the probability of not rejecting the null hypothesis when it is false (i.e., the type II error) depends on the chosen significance level, sample size, and effect size. Without knowing the effect size, it is not possible to calculate the type II error.

(b) To calculate the required sample size to detect a true mean diameter of 1.55 inches with a power of at least 0.9, we need to know the chosen significance level, the standard deviation of the population, and the effect size.

Using a statistical power calculator or a sample size formula, we can determine that a sample size of approximately 65 bearings is needed.

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1 hectare is defined as 1 x 10^4 m^2. 1 acre is 4.356 x 10^4 ft. How many acres are in 2.0 hectares? (Do not include units in your answer).

Answers

There are approximately 0.4594 acres in 2.0 hectares.

To solve this problem

We need to use the conversion factor between hectares and acres.

Given:

[tex]1 hectare = 1[/tex] × [tex]10^4 m^2[/tex]

[tex]1 acre = 4.356[/tex] × [tex]10^4 ft[/tex]

To find the number of acres in 2.0 hectares, we can set up the following conversion:

[tex]2.0 hectares * (1[/tex] × [tex]10^4 m^2 / 1 hectare) * (1 acre / 4.356[/tex] × [tex]10^4 ft)[/tex]

Simplifying the units:

[tex]2.0 * (1[/tex] × [tex]10^4 m^2) * (1 acre / 4.356[/tex] ×[tex]10^4 ft)[/tex]

Now, we can perform the calculation:

[tex]2.0 * (1[/tex] × [tex]10^4) * (1 /[/tex][tex]4.356[/tex] ×[tex]10^4)[/tex]

= 2.0 * 1 / 4.356

= 0.4594

Therefore, there are approximately 0.4594 acres in 2.0 hectares.

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In triangle ABC the angle bisectors drawn from vertices A and B intersect at point D. Find m
m

Answers

The measure of angle ADB is equal to the square root of ([tex]AB \times BA[/tex]).

In triangle ABC, let the angle bisectors drawn from vertices A and B intersect at point D. To find the measure of angle ADB, we can use the angle bisector theorem. According to this theorem, the angle bisector divides the opposite side in the ratio of the adjacent sides.

Let AD and BD intersect side BC at points E and F, respectively. Now, we have triangle ADE and triangle BDF.

Using the angle bisector theorem in triangle ADE, we can write:

AE/ED = AB/BD

Similarly, in triangle BDF, we have:

BF/FD = BA/AD

Since both angles ADB and ADF share the same side AD, we can combine the above equations to obtain:

(AE/ED) * (FD/BF) = (AB/BD) * (BA/AD)

By substituting the given angle bisector ratios and rearranging, we get:

(AD/BD) * (AD/BD) = (AB/BD) * (BA/AD)

AD^2 = AB * BA

Note: The solution provided assumes that points A, B, and C are non-collinear and that the triangle is non-degenerate.

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Determine the first three nonzero terms in the Taylor polynomial approximation for the given initial value problem. x ′′
+8tx=0;x(0)=1,x ′
(0)=0 The Taylor approximation to three nonzero terms is x(t)=+⋯.

Answers

The first three nonzero terms in the Taylor polynomial approximation for the given initial value problem are: 1 - t^2/8 + t^4/128.

Given the initial value problem: x′′ + 8tx = 0; x(0) = 1, x′(0) = 0. To find the first three nonzero terms in the Taylor polynomial approximation, we follow these steps:

Step 1: Find x(t) and x′(t) using the integrating factor.

We start with the differential equation x′′ + 8tx = 0. Taking the integrating factor as I.F = e^∫8t dt = e^4t, we multiply it on both sides of the equation to get e^4tx′′ + 8te^4tx = 0. This simplifies to e^4tx′′ + d/dt(e^4tx') = 0.

Integrating both sides gives us ∫ e^4tx′′ dt + ∫ d/dt(e^4tx') dt = c1. Now, we have e^4tx' = c2. Differentiating both sides with respect to t, we get 4e^4tx' + e^4tx′′ = 0. Substituting the value of e^4tx′′ in the previous equation, we have -4e^4tx' + d/dt(e^4tx') = 0.

Simplifying further, we get -4x′ + x″ = 0, which leads to x(t) = c3e^(4t) + c4.

Step 2: Determine the values of c3 and c4 using the initial conditions.

Using the initial conditions x(0) = 1 and x′(0) = 0, we can substitute these values into the expression for x(t). This gives us c3 = 1 and c4 = -1/4.

Step 3: Write the Taylor polynomial approximation.

The Taylor approximation to three nonzero terms is x(t) = 1 - t^2/8 + t^4/128 + ...

Therefore, the starting value problem's Taylor polynomial approximation's first three nonzero terms are: 1 - t^2/8 + t^4/128.

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Jocelyn estimates that a piece of wood measures 5.5 cm. If it actually measures 5.62 cm, what is the percent error of Jocelyn’s estimate?

Answers

Answer:

The percent error is -2.1352% of Jocelyn's estimate.

In the figure, the square ABCD and the AABE are standing on the same base AB and between the same parallel lines AB and DE. If BD = 6 cm, find the area of AEB.​

Answers

To find the area of triangle AEB, we use base AB (6 cm) and height 6 cm. Applying the formula (1/2) * base * height, the area is 18 cm².

To find the area of triangle AEB, we need to determine the length of the base AB and the height of the triangle. Since both square ABCD and triangle AABE is standing on the same base AB, the length of AB remains the same for both.

We are given that BD = 6 cm, which means that the length of AB is also 6 cm. Now, to find the height of the triangle, we can consider the height of the square. Since AB is the base of both the square and the triangle, the height of the square is equal to AB.

Therefore, the height of triangle AEB is also 6 cm. Now we can calculate the area of the triangle using the formula: Area = (1/2) * base * height. Plugging in the values, we get Area = (1/2) * 6 cm * 6 cm = 18 cm².

Thus, the area of triangle AEB is 18 square centimeters.

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Record the following information below. Be sure to clearly notate which number is which parameter. A.) time of five rotations B.) time of one rotation C.) distance from the shoulder to the elbow D.) distance from the shoulder to the middle of the hand. A. What was the average angular speed (degrees/s and rad/s) of the hand? B. What was the average linear speed (m/s) of the hand? C. Are the answers to A and B the same or different? Explain your answer.

Answers

The average angular speed of the hand is ω = 1800 / t rad/s and 103140 / t degrees/s and the average linear speed of the hand is 5D / t m/s.  The answers to A and B are not the same as they refer to different quantities with different units and different values.

A) To find the average angular speed of the hand, we need to use the formula:

angular speed (ω) = (angular displacement (θ) /time taken(t))

= 5 × 360 / t

Here, t is the time for 5 rotations

So, average angular speed of the hand is ω = 1800 / trad/s

To convert this into degrees/s, we can use the conversion:

1 rad/s = 57.3 degrees/s

Therefore, ω in degrees/s = (ω in rad/s) × 57.3

= (1800 / t) × 57.3

= 103140 / t degrees/s

B) To find the average linear speed of the hand, we need to use the formula:linear speed (v) = distance (d) /time taken(t)

Here, the distance of the hand is the length of the arm.

Distance from shoulder to middle of hand = D

Similarly, the time taken to complete 5 rotations is t

Thus, the total distance covered by the hand in 5 rotations is D × 5

Therefore, average linear speed of the hand = (D × 5) / t

= 5D / t

= 5 × distance of hand / time for 5 rotations

C) No, the answers to A and B are not the same. This is because angular speed and linear speed are different quantities. Angular speed refers to the rate of change of angular displacement with respect to time whereas linear speed refers to the rate of change of linear displacement with respect to time. Therefore, they have different units and different values.

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