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Allisa [31]
4 years ago
13

What's the lenght of side b in the figure below

Mathematics
1 answer:
Evgen [1.6K]4 years ago
3 0
Let's use Pythagorean theorem.  a^{2}+b^{2}=c^{2} where c is the hypotenuse.

So if we set it up:

7^{2}+b^{2}=13^{2}

now we simplify:

49+b^{2}=169

and get b^{2} by itself:

b^{2}=120

Now to get b by itself, we need to take the square root of both sides:

b=\sqrt{120}

And that's your answer!
You might be interested in
Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +
Veronika [31]

Answer:

Verified

y(x) = \frac{3Ln(x) + 3}{x}

y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)

- Therefore, the complete solution to the given ODE can be expressed as:

                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

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6 0
3 years ago
Need help <br> Choices<br> 4<br> 3<br> 2<br> 1
-Dominant- [34]

Answer:

1

Step-by-step explanation:

The amplitude is the maximum distance away from the middle of the wave. Here we can see that the middle of the wave is the x axis and the farthest point (largest difference between the y coordinate of the x-axis, or the line y = 0, and the wave) is 1 unit away.

4 0
3 years ago
Read 2 more answers
1/2-1/3 need help with this problem
jeka57 [31]

Answer: 1/6

Step-by-step explanation:

Common denominator is 6 the 1/2 is. 3/6

the 1/3 is 2/6

Subtract. You get 1/6

4 0
3 years ago
Read 2 more answers
Please help a friend
choli [55]

Answer:

C.

Step-by-step explanation:

8 0
3 years ago
The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th
inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4   -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

y(y-2)=100

y^2-2y=100

subtracting 100 from both sides

y^2-2y-100=0

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

ax^2+bx+c=0

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}

y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}

Solving first

y=\frac{2+\sqrt{4+400}{2}

y=\frac{2+\sqrt{404}{2}

y=\frac{2+20.099}{2}

y=\frac{22.099}{2}

y=11.049

Solving second one

y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}

y=\frac{2-\sqrt{4+400}{2}

y=\frac{2-\sqrt{404}{2}

y=\frac{2-20.099}{2}

y=\frac{-18.99}{2}

y=-9.045

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

4 0
3 years ago
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