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3 years ago

PLS HELP ASAP WILL GIVE POINTS AND BRAINLIEST THANK YOU.

Mathematics

1 answer:

kifflom [539]3 years ago

4 0

this is an awkward gather pic, that guy with the blue sweatshirt reminds me of a friend from my school. Although what you should do is try to give people more space for people to fit in right in the picture. Maybe like for the short kids should go to the front and the talk kids go to the back. This is just scram.bled

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2 Solve the equation 3x +125 = 49

Alex787 [66]

Qustion:

3x +125 = 49

Answer:

x = - 25 1/3

8 0

2 years ago

Please help me!!!!!

denpristay [2]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → A = π - (B + C)

→ B = π - (A + C)

→ C = π - (A + B)

Use Sum to Product Identity: sin A - sin B = 2 cos [(A + B)/2] · sin [(A - B)/2]

Use the following Cofunction Identity: cos (π/2 - A) = sin A

Proof LHS → RHS:

LHS: sin A - sin B + sin C

= (sin A - sin B) + sin C

$\text{Sum to Product:}\quad 2\cos \bigg(\dfrac{A+B}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

$\text{Given:}\qquad 2\cos \bigg(\dfrac{\pi -(B+C)}{2}+\dfrac{B}{2}}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad =2\cos \bigg(\dfrac{\pi -C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

$.\qquad \qquad =2\cos \bigg(\dfrac{\pi}{2} -\dfrac{C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

$\text{Cofunction:} \qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)+2\cos \bigg(\dfrac{C}2{}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

$\text{Factor:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\bigg]$

$\text{Given:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)\bigg]\\\\\\.\qquad \qquad =2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi}{2} -\dfrac{(A+B)}{2}\bigg)\bigg]$

$\text{Cofunction:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)\bigg]$

$\text{Sum to Product:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ 2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad \qquad =4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)$

$\text{LHS = RHS:}\quad 4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)=4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)\quad \checkmark$

6 0

2 years ago

Solve each triangle. Round your answers to the nearest tenth

ivolga24 [154]

You have to use the Law of Cosines here, since there's no other way to solve this. it's not a right triangle, so you can't use the Pythagorean Theorem. The Law of Cosines will help us find the missing side length then we will have to use the Law of Sines to find another angle. Then after that we will use the Triangle Angle-Sum theorem to finish it off. Ready? The Law of Cosines to find side b is $b^{2} = a^{2}+ c^{2} -2ac cosB$ and fill in the info we know, which is everything but the b. $b^{2} =(21) ^{2} +(29) ^{2} -2(21)(29)cos109$ . Doing all that math gives us that side b = 40.9 or 41. Now the Law of Sines to find missing angle A or C. Let's find A. $\frac{sinA}{21} = \frac{sin109}{41}$ . That gives us that angle A is 29. Now use the fact that all triangles add up to 180 to get that angle C is 42. And you're done!

7 0

3 years ago

Pls help will give brainiest

Shalnov [3]

Answer:

0.75

Step-by-step explanation:

5 0

2 years ago

Your price on a particular model is $125 however to get a service contract you offer to sell it for $115 how much discount are y

Arisa [49]

It would be an 8 percent discount because 115 is 92 percent of 125.

6 0

3 years ago