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Strike441 [17]
3 years ago
15

Prove that the angle sum of any triangle is 180°

Mathematics
1 answer:
Licemer1 [7]3 years ago
3 0
I will be philosophical here...

A square is 4 -90degree angles (definition of perpendicular lines)

If we create a diagonal it splits it into two triangles which in result gives us

90degree + 90/2 degree+ 90/2 degree (because the diagonal split it in half and one full 90 by definition of perpendicular lines)

That gives us the beautiful 180 degrees

which is also what humans use to say that they’ve completely changed or did something completely different...

Mysteries of the triangle...

Six six six six six six
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jennifer scored 75 on her quiz last week. this week she scored 90. what was the percent of increase on the quiz score?
Arlecino [84]

Answer:

The answer is a 20% increase

Hope this helps!

Mark me brainliest if I'm right :)

4 0
3 years ago
Please help me with this.
svlad2 [7]

The correct answer is C.) (y|y > -4).

In order to find the range, we look for the possible numbers that y can be. In this case, you'll see the lowest the number goes is -4 and it continues up forever. Therefore, the answer is that y has to be greater than -4.

6 0
3 years ago
Helppppppppppppppppp
Maksim231197 [3]

\huge \tt༆ Answer ༄

The respective terms used for the lines are ~

  • a.) Line a = Tangent

  • b.) Line b = Radius

  • c.) Line c = Diameter

  • d.) Line d = Secant
3 0
3 years ago
Read 2 more answers
S(t)=9t-4, find S(4)
melamori03 [73]
For this problem, you need to substitute 4 in for your variable t. the original equation is S(t) and the one you need is S(4) -- in other words, evaluate this equation when t = 4.

S(4) = 9(4) - 4
S(4) = 36 - 4
S(4) = 32 is your answer.
3 0
4 years ago
D/d{cosec^-1(1+x²/2x)} is equal to​
SIZIF [17.4K]

Step-by-step explanation:

\large\underline{\sf{Solution-}}

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

Let assume that

\rm :\longmapsto\:y =  {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

We know,

\boxed{\tt{  {cosec}^{ - 1}x =  {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}

So, using this, we get

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2x}{1 +  {x}^{2} } \bigg)

Now, we use Method of Substitution, So we substitute

\red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z =  {tan}^{ - 1}x}

So, above expression can be rewritten as

\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 +  {tan}^{2} z} \bigg)

\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)

\rm\implies \:y = 2z

\bf\implies \:y = 2 {tan}^{ - 1}x

So,

\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x

Thus,

\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg)

\rm \:  =  \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)

\rm \:  =  \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)

\rm \:  =  \: 2 \times \dfrac{1}{1 +  {x}^{2} }

\rm \:  =  \: \dfrac{2}{1 +  {x}^{2} }

<u>Hence, </u>

\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 +  {x}^{2} }{2x} \bigg) =  \frac{2}{1 +  {x}^{2} }}}}

<u>Hence, Option (d) is </u><u>correct.</u>

6 0
2 years ago
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