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lesantik [10]
3 years ago
10

Is there any tricks to memorize the special right triangle's sides?

Mathematics
2 answers:
noname [10]3 years ago
8 0
Sohcahtoa
My algebra teacher got us all to remember that word because...
Soh - Sin = Opposite / Hypotenus
Cah - Cos = Adjacent / Hypotenus
Toa - Tan = Opposite / Adjacent
Sin, Cos, and Tan are what you can use to find missing triangle sides if you know at least one other angle besides the 90 degree angle.

Zolol [24]3 years ago
5 0
No, but there is a way of creating them. If you let 
a = 2xy
b = x^2 - y^2 where x > y
c = x^2 + y^2 

And a^2 + b^2 = c^2 You can create these sides to your needs. For example

x = 5
y = 2

2xy = 2*5*2 = 20
x^2 - y^2 = 25 - 4 = 21
x^2 + y^2 = 25 + 4 = 29  which is not well known but it is a right angle.

29^2 = 20^2 + 21^2

841 = 400 + 441
841 = 841

Just one further comment. It is easy to check to see if the triangle you are talking about is a right triangle. Just use a^2 + b^2 = c^2. 

Generally it is unnecessary to either create your own or memorize ones you have come across. 

I think any math teacher would say "why bother?"
If you are not satisfied, put a reason for doing this in the comments. 
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What is the median,mode, and mean of 76,76,77,84,85,89
bogdanovich [222]
76 is the mean for this problem
4 0
2 years ago
On a coordinate plane, kite W X Y Z is shown. Point W is at (negative 3, 3), point X is at (2, 3), point Y is at (4, negative 4)
xz_007 [3.2K]

Answer:

P = 10 + 2\sqrt{53} units

Step-by-step explanation:

Given

Shape: Kite WXYZ

W (-3, 3),  X (2, 3),

Y (4, -4),  Z (-3, -2)

Required

Determine perimeter of the kite

First, we need to determine lengths of sides WX, XY, YZ and ZW using distance formula;

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

For WX:

(x_1, y_1)\ (x_2,y_2) = (-3, 3),\ (2, 3)

WX = \sqrt{(-3 - 2)^2 + (3 - 3)^2}

WX = \sqrt{(-5)^2 + (0)^2}

WX = \sqrt{25}

WX = 5

For XY:

(x_1, y_1)\ (x_2,y_2) = (2, 3)\ (4,-4)

XY = \sqrt{(2 - 4)^2 + (3 - (-4))^2}

XY = \sqrt{-2^2 + (3 +4)^2}

XY = \sqrt{-2^2 + 7^2}

XY = \sqrt{4 + 49}

XY = \sqrt{53}

For YZ:

(x_1, y_1)\ (x_2,y_2) = (4,-4)\ (-3, -2)

YZ = \sqrt{(4 - (-3))^2 + (-4 - (-2))^2}

YZ = \sqrt{(4 +3)^2 + (-4 +2)^2}

YZ = \sqrt{7^2 + (-2)^2}

YZ = \sqrt{49 + 4}

YZ = \sqrt{53}

For ZW:

(x_1, y_1)\ (x_2,y_2) = (-3, -2)\ (-3, 3)

ZW = \sqrt{(-3 - (-3))^2 + (-2 - 3)^2}

ZW = \sqrt{(-3 +3)^2 + (-2 - 3)^2}

ZW = \sqrt{0^2 + (-5)^2}

ZW = \sqrt{0 + 25}

ZW = \sqrt{25}

ZW = 5

The Perimeter (P) is as follows:

P = WX + XY + YZ + ZW

P = 5 + \sqrt{53} + \sqrt{53} + 5

P = 5 + 5 + \sqrt{53} + \sqrt{53}

P = 10 + 2\sqrt{53} units

6 0
3 years ago
Ragon, who lives on this planet, does a survey and finds that her colony of 237 contains 45 black-footed, one-headed Mizjigs; 69
Gemiola [76]

The answer is <u>A. 98</u>.

Because when you subtract 115 - 45 it equals 70 then you add 70 + 69 which equals 139 then you subtract 237 - 139 it equals <u>98</u> which is your answer!

<h2>      Hope this helps!</h2>
4 0
3 years ago
Sin75degree=Cos____degree
tigry1 [53]

Answer: 15

Step-by-step explanation:

We know that sin 75 degrees = 0.90659...

Now, to find cos: cos-1 (0.90659) = 15

So the answer is 15.

7 0
2 years ago
Pleaseeeee ı dont find the answer
mote1985 [20]

Answer:  y(x) = \sqrt{\frac{7x^{14}}{-2x^7+9}}\\\\

==========================================================

Explanation:

The given differential equation (DE) is

y'-\frac{7}{x}y = \frac{y^3}{x^8}\\\\

Which is the same as

y'-\frac{7}{x}y = \frac{1}{x^8}y^3\\\\

This 2nd DE is in the form y' + P(x)y = Q(x)y^n

where

P(x) = -\frac{7}{x}\\\\Q(x) = \frac{1}{x^8}\\\\n = 3

As the instructions state, we'll use the substitution u = y^{1-n}

We specifically use u = y^{1-n} = y^{1-3} = y^{-2}

-----------------

After making the substitution, we'll end up with this form

\frac{du}{dx} + (1-n)P(x)u = (1-n)Q(x)\\\\

Plugging in the items mentioned, we get:

\frac{du}{dx} + (1-n)P(x)u = (1-n)Q(x)\\\\\frac{du}{dx} + (1-3)*\frac{-7}{x}u = (1-3)\frac{1}{x^8}\\\\\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\

We can see that we have a new P(x) and Q(x)

P(x) = \frac{14}{x}\\\\Q(x) = -\frac{2}{x^8}

-------------------

To solve the linear DE \frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\, we'll need the integrating factor which I'll call m

m(x) = e^{\int P(x) dx} = e^{\int \frac{14}{x}dx} = e^{14\ln(x)}

m(x) = e^{\ln(x^{14})} = x^{14}

We will multiply both sides of the linear DE by this m(x) integrating factor to help with further integration down the road.

\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\m(x)*\left(\frac{du}{dx} + \frac{14}{x}u\right) = m(x)*\left(-\frac{2}{x^8}\right)\\\\x^{14}*\frac{du}{dx} + x^{14}*\frac{14}{x}u = x^{14}*\left(-\frac{2}{x^8}\right)\\\\x^{14}*\frac{du}{dx} + 14x^{13}*u = -2x^6\\\\\left(x^{14}*u\right)' = -2x^6\\\\

It might help to think of the product rule being done in reverse.

Now we can integrate both sides to solve for u

\left(x^{14}*u\right)' = -2x^6\\\\\displaystyle \int\left(x^{14}*u\right)'dx = \int -2x^6 dx\\\\\displaystyle x^{14}*u = \frac{-2x^7}{7}+C\\\\\displaystyle u = x^{-14}*\left(\frac{-2x^7}{7}+C\right)\\\\\displaystyle u = x^{-14}*\frac{-2x^7}{7}+Cx^{-14}\\\\\displaystyle u = \frac{-2x^{-7}}{7}+Cx^{-14}\\\\

u = \frac{-2}{7x^7} + \frac{C}{x^{14}}\\\\u = \frac{-2}{7x^7}*\frac{x^7}{x^7} + \frac{C}{x^{14}}*\frac{7}{7}\\\\u = \frac{-2x^7}{7x^{14}} + \frac{7C}{7x^{14}}\\\\u = \frac{-2x^7+7C}{7x^{14}}\\\\

Unfortunately, this isn't the last step. We still need to find y.

Recall that we found u = y^{-2}\\\\

So,

u = \frac{-2x^7+7C}{7x^{14}}\\\\y^{-2} = \frac{-2x^7+7C}{7x^{14}}\\\\y^{2} = \frac{7x^{14}}{-2x^7+7C}

We're told that y(1) = 1. This means plugging x = 1 leads to the output y = 1. So the RHS of the last equation should lead to 1. We'll plug x = 1 into that RHS, set the result equal to 1 and solve for C

\frac{7*1^{14}}{-2*1^7+7C} = 1\\\\\frac{7}{-2+7C} = 1\\\\7 = -2+7C\\\\7+2 = 7C\\\\7C = 9\\\\C = \frac{9}{7}

So,

y^{2} = \frac{7x^{14}}{-2x^7+7C}\\\\y^{2} = \frac{7x^{14}}{-2x^7+7*\frac{9}{7}}\\\\y^{2} = \frac{7x^{14}}{-2x^7+9}\\\\y = \sqrt{\frac{7x^{14}}{-2x^7+9}}\\\\

We go with the positive version of the root because y(1) is positive, which must mean y(x) is positive for all x in the domain.

3 0
2 years ago
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