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3 years ago

PLEASE HELP ME GUYS ×0×Find the RATIO and the EXACT VALUE of the given Tan A.

Mathematics

1 answer:

ikadub [295]3 years ago

5 0

Answer:

Step-by-step explanation:

take angle A as reference angle

using tan rule

tan A=opposite/adjacent

tan A=5/12

tan A=0.41

A=tan 22.5

A=22.5

for ratio,

tan A=5/12

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4vir4ik [10]

Answer:

Step-by-step explanation:

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5 0

3 years ago

A direct variation exist between x and y. When x=4, y=3. What is the value of y when x=10

scoray [572]

Answer:

$y = 7.5$

Step-by-step explanation:

Given

Direct variation implies that:

$y = kx$

Where k is the constant of variation.

$x=4; y=3$

Required

Find y when $x = 10$

$y = kx$

Make k the subject

$k = \frac{y}{x}$

For: $x=4; y=3$

$k = \frac{3}{4}$

For: $x = 10$

$k = \frac{y}{x}$ becomes

$k = \frac{y}{10}$

Substitute $k = \frac{3}{4}$

$\frac{3}{4} = \frac{y}{10}$

Make y the subject

$y = \frac{3}{4} * 10$

$y = \frac{3* 10}{4}$

$y = \frac{30}{4}$

$y = 7.5$

7 0

4 years ago

1. Which shows another way to write 73?

7nadin3 [17]

1. The answer is C. 7 with an exponent of 3 means that you multiply 7 by itself three times, or 7 x 7 x 7.
2. The answer is A. 19 with an exponent of 3 is the same as 19 x 19 x 19.

6 0

4 years ago

Solve the differential equation dy/dx=x/49y. Find an implicit solution and put your answer in the following form: = constant. he

anygoal [31]

Answer:

The general solution of the differential equation is $\frac{49y^{2} }{2}-\frac{x^{2} }{2} = c_{3}$

The equation of the solution through the point (x,y)=(7,1) is $y=\frac{x}{7}$

The equation of the solution through the point (x,y)=(0,-3) is $\:y=-\frac{\sqrt{441+x^2}}{7}$

Step-by-step explanation:

This differential equation $\frac{dy}{dx}=\frac{x}{49y}$ is a separable first-order differential equation.

We know this because a first order differential equation (ODE) $y' =f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of x and y

$f(x,y)=p(x)\cdot h(y)$ where p(x) and h(y) are continuous functions. And this ODE is equal to $\frac{dy}{dx}=x\cdot \frac{1}{49y}$

To solve this differential equation we rewrite in this form:

$49y\cdot dy=x \cdot dx$

And next we integrate both sides

$\int\limits {49y} \, dy=\int\limits {x} \, dx$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1}\\\int\limits {49y} \, dy=\frac{49y^{2} }{2} + c_{1}$

$\int\limits {x} \, dx=\frac{x^{2} }{2} +c_{2}$

$\int\limits {49y} \, dy=\int\limits {x} \, dx\\\frac{49y^{2} }{2} + c_{1} =\frac{x^{2} }{2} +c_{2}$

We can subtract constants $c_{3}=c_{2}-c_{1}$

$\frac{49y^{2} }{2} =\frac{x^{2} }{2} +c_{3}$

An explicit solution is any solution that is given in the form $y=y(t)$ . That means that the only place that y actually shows up is once on the left side and only raised to the first power.

An implicit solution is any solution of the form $f(x,y)=g(x,y)$ which means that y and x are mixed (y is not expressed in terms of x only).

The general solution of this differential equation is:

$\frac{49y^{2} }{2}-\frac{x^{2} }{2} = c_{3}$

To find the equation of the solution through the point (x,y)=(7,1)

We find the value of the $c_{3}$ with the help of the point (x,y)=(7,1)

$\frac{49*1^2\:}{2}-\frac{7^2\:}{2}\:=\:c_3\\c_3 = 0$

Plug this into the general solution and then solve to get an explicit solution.

$\frac{49y^2\:}{2}-\frac{x^2\:}{2}\:=\:0$

$\mathrm{Add\:}\frac{x^2}{2}\mathrm{\:to\:both\:sides}\\\frac{49y^2}{2}-\frac{x^2}{2}+\frac{x^2}{2}=0+\frac{x^2}{2}\\Simplify\\\frac{49y^2}{2}=\frac{x^2}{2}\\\mathrm{Multiply\:both\:sides\:by\:}2\\\frac{2\cdot \:49y^2}{2}=\frac{2x^2}{2}\\Simplify\\9y^2=x^2\\\mathrm{Divide\:both\:sides\:by\:}49\\\frac{49y^2}{49}=\frac{x^2}{49}\\Simplify\\y^2=\frac{x^2}{49}\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$y=\frac{x}{7},\:y=-\frac{x}{7}$

We need to check the solutions by applying the initial conditions

With the first solution we get:

$y=\frac{x}{7}=\\1=\frac{7}{7}\\1=1\\$

With the second solution we get:

$\:y=-\frac{x}{7}\\1=-\frac{7}{7}\\1\neq -1$

Therefore the equation of the solution through the point (x,y)=(7,1) is $y=\frac{x}{7}$

To find the equation of the solution through the point (x,y)=(0,-3)

We find the value of the $c_{3}$ with the help of the point (x,y)=(0,-3)

$\frac{49*-3^2\:}{2}-\frac{0^2\:}{2}\:=\:c_3\\c_3 = \frac{441}{2}$

Plug this into the general solution and then solve to get an explicit solution.

$\frac{49y^2\:}{2}-\frac{x^2\:}{2}\:=\:\frac{441}{2}$

$y^2=\frac{441+x^2}{49}\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\y=\frac{\sqrt{441+x^2}}{7},\:y=-\frac{\sqrt{441+x^2}}{7}$

We need to check the solutions by applying the initial conditions

With the first solution we get:

$y=\frac{\sqrt{441+x^2}}{7}\\-3=\frac{\sqrt{441+0^2}}{7}\\-3\neq 3$

With the second solution we get:

$y=-\frac{\sqrt{441+x^2}}{7}\\-3=-\frac{\sqrt{441+0^2}}{7}\\-3=-3$

Therefore the equation of the solution through the point (x,y)=(0,-3) is $\:y=-\frac{\sqrt{441+x^2}}{7}$

4 0

4 years ago

Find the lengths of AB and AD A(0,-6)B(-3,0), C(4,4), D(7,-2)

12345 [234]

Answer: The lengths between AB and AD is 0.

Step-by-step explanation:

(0,6) Find the difference in Y and the difference in x .

(-3,0)

3^2 + 6 ^2 = Length^2

9 + 36= length^2

45 = Length^2

(4,4)

(7,-2)

3^2 + 6^2= length^2

9 + 36 = length^2

45= length^2

8 0

3 years ago

PLEASE HELP ME GUYS ×0×Find the RATIO and the EXACT VALUE of the given Tan A.​

PLEASE HELP ME GUYS ×0×Find the RATIO and the EXACT VALUE of the given Tan A.