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2 years ago

ALGEBRA Identify the similar triangles. Then find each measure.

Mathematics

1 answer:

Elena L [17]2 years ago

3 0

Answer:

fg t f h 10

Step-by-step explanation:

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Can someone please help me it’s very important please ASAP

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The answer to this question is c)

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2 years ago

Hey can some one help me out

Archy [21]

I believe the correct situation is C

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2 years ago

A crew of highway workers paved 1/5 mile in 1/8 hour. If they work at the same rate, how much will they

Elan Coil [88]

Answer:

The easiest approach is to realise that one hour is 3 times longer than 20 minutes. The longer the time, the more they will pave.

215 of a mile, in 20 minutes, how much in 60 minutes?#

They will pave 3 times more.

215×31=615 of a mile

615=25 of a mile

You could also use the 'unitary method' where you find out how much they pave in ONE minute (divide by 20) and them multiply by 60 to find how much in one hour.

Look at what happens:

215÷20×60

=215×120×603

=215×3 ← exactly the same maths.

=25

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2 years ago

[please help fast]

madreJ [45]

SAS because of the side, angle and the reflexive which is a side.

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3 years ago

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

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