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seraphim [82]
3 years ago
6

What is the longest side of a right triangle called?

Mathematics
2 answers:
liraira [26]3 years ago
5 0

Answer:

Step-by-step explanation:

hypotenuse

Nadya [2.5K]3 years ago
5 0
The hypotenuse is always opposite the right angle and it is always the longest side of the triangle - famously called as “Pythagoras’ Theorem”
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WILL AWARD BRAINLIEST ANSWER
Tpy6a [65]

Answer:

(x - 5)^2 + (y - 7)^2 = 13^2

or

(x - 5)^2 + (y - 7)^2 = 169

Step-by-step explanation:

Well, the center origin of the circle is given (h,k) =  (5,7).

We have to find our radius as they gave us a point. from origin to the edge of the circle.

Using the formula: (x - h)^2 + (y - k)^2 = r^2

Plug in our (h,k) = (5,7) and (x,y) =  (10,19) to solve for radius.

(x - h)^2 + (y - k)^2 = r^2

(10 - (5))^2 + (19 - (7)^2 = r^2

(5)^2 + (12)^2 = r^2

25 + 144  = r^2

r^2 = 169

r = 13

5 0
2 years ago
Activity 4: Carnival Tickets
Vilka [71]

Answer:

100 kids tickets where sold

Step-by-step explanation:

8 0
4 years ago
Write your answer as a fraction or whole number without exponents. 3^–1 plz help i have no idea how to do this
Thepotemich [5.8K]

Answer:

1/3

Step-by-step explanation:

Hope i helped you

7 0
3 years ago
Read 2 more answers
Simplify tan9x-tan5x / 1+tan9xtan5x<br> (SHOW WORK)
Charra [1.4K]

Answer:

The simplest form is tan(4x)

Step-by-step explanation:

* Lets revise the identity of the compound angles

- tan(a+b)=\frac{tan(a)+tan(b)}{1-tan(a)tan(b)}

- tan(a-b)=\frac{tan(a)-tan(b)}{1+tan(a)tan(b)}

* Lets solve the problem

- Let 9x = 5x + 4x

∴ tan(9x) = tan(5x + 4x)

- Use the rule of the compound angle

∵ \frac{tan(9x)-tan(5x)}{1+tan(9x)tan(5x)} ⇒ (1)

∵ tan(5x+4x)=\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)} ⇒ (2)

∵ tan(9x) = equation (2)

- Substitute (2) in (1)

∴ \frac{\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)}-tan(5x)}{1+(\frac{tan(5x)+tan(4x)}{1-tan(5x)tan(4x)})tan(5x)}

- Multiply up and down by (1 - tan(5x)tan(4x))

∴ \frac{tan(5x)+tan(4x)-tan(5x)[1-tan(5x)tan(4x)]}{1-tan(5x)tan(4x)+tan(5x)[tan(5x)+tan(4x)]}

- Simplify up and down

∴ \frac{tan(5x)+tan(4x)-tan(5x)+tan^{2}(5x)tan(4x)}{1-tan(5x)tan(4x)+tan^{2}(5x)+tan(5x)tan(4x) }

∴ \frac{tan(4x)+tan^{2}(5x)tan(4x)}{[1+tan^{2}(5x)]}

- Take tan(4x) as a common factor up

∴ \frac{tan(4x)[1+tan^{2}(5x)]}{[1+tan^{2}(5x)]}

- Cancel [1 + tan²(5x)] up and down

∴ The answer is tan(4x)

4 0
4 years ago
Do you always copy change change while subtracting negative integers?
Evgesh-ka [11]
Think of it like this: imagine a frog on some lily pads and if it hops left it gets smaller(negative) and if it hops right it gets bigger(positive) ,(but think of it like numbers) so -14-5 think about the -14 being way down at the left and then you add on the (-5) since the number you are adding is negative it is getting smaller (going farther away from the zero, not closer) so when you add a negative to a negative number you get smaller instead of bigger. Hope that helps
3 0
3 years ago
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