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bulgar [2K]
3 years ago
11

Is it (sometimes true/never true) that all three altitudes of a triangle fall outside the triangle?

Mathematics
1 answer:
aniked [119]3 years ago
5 0

never true is always 180

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Please help me finish these for summer school :)
ValentinkaMS [17]
13.45 will be the answer

since we are finding the hypotenuse, the formula is a^2 + b^2 = c^2

this will then be 10^2 + 9^2 = c^2

181=c^2

c=13.45 :)
5 0
3 years ago
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A mixture of alcohol and water contains a total of 50 ounces of liquid. There are 10 ounces of pure alcohol in the mixture. What
suter [353]
20% is alcohol and 80% is water hope this helped
5 0
3 years ago
The area of a rectangle or a 48 square inches the length is 8 inches what is the measure of its width
adoni [48]
6 inches because area= length times width and 6 times 8= 48 inches
5 0
3 years ago
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What is the value of E(-1) n (3n+2)?
RoseWind [281]
<h3>Answer:  C) 6</h3>

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

Explanation:

The weird looking E symbol is the greek uppercase letter sigma. It refers to a sum.

It tells us to add up terms in the form (-1)^n*(3n+2) where n is an integer ranging from n = 1 to n = 4.

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

If n = 1, then we have

(-1)^n*(3n+2) = (-1)^1*(3*1+2) = -5

Let A = -5 as we'll use it later.

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

If n = 2, then

(-1)^n*(3n+2) = (-1)^2*(3*2+2) = 8

Let B = 8 since we'll use this later as well

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

If n = 3, then

(-1)^n*(3n+2) = (-1)^3*(3*3+2) = -11

Let C = -11

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

If n = 4, then

(-1)^n*(3n+2) = (-1)^4*(3*4+2) = 14

Let D = 14.

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

We'll add up the values of A,B,C,D to get the final answer

A+B+C+D = -5+8+(-11)+14 = 6

This means that

\displaystyle \sum_{n=1}^{4}(-1)^n(3n+2) = 6

6 0
3 years ago
Simplify the expression using trigonometric identities: sec (–θ) – cos θ.
ololo11 [35]

Answer:

sin θ . tan θ

Step-by-step explanation:

Note : -

sec ( - θ ) = sec θ

Formula / Identity : -

sec θ = 1 / cos θ

sec ( - θ ) - cos θ

= [ 1 / cos θ ] - cos θ

{ LCM = cos θ }

= [ 1 / cos θ ] - [ cos²θ / cos θ ]

= [ 1 - cos²θ ] / cos θ

{ 1 - cos²θ = sin²θ }

= sin²θ / cos θ

{ sin²θ = sin θ . sin θ }

= sin θ . sin θ / cos θ

{ sin θ / cos θ = tan θ }

= sin θ . tan θ

Hence, simplified.

4 0
3 years ago
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