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3 years ago

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PLEASE HELP!! can you solve this problem? you need to fill all 10 squares with the correct number and/or symbols. Below are the

options of numbers and/or symbols that can be placed in the grid. And above the grid is the problem!!

1 answer:

Aliun [14]3 years ago

8 0

Answer:

Wow just wow

Step-by-step explanation:

I have no clue sorry ;-;

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Please help me!!!!!

denpristay [2]

Answer: see proof below

<u>Step-by-step explanation:</u>

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

<u>Proof LHS → RHS:</u>

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

3 years ago

For a given population of high school seniors, the Scholastic Aptitude Test (SAT) in mathematics has a mean score of 500 with a

tresset_1 [31]

Answer:

The probability that a randomly selected high school senior's score on mathematics part of SAT will be

(a) more than 675 is 0.0401

(b)between 450 and 675 is 0.6514

Step-by-step explanation:

Mean of Sat = $\mu = 500$

Standard deviation = $\sigma = 100$

We will use z score over here

What is the probability that a randomly selected high school senior's score on mathe- matics part of SAT will be

(a) more than 675?

P(X>675)

$Z=\frac{x-\mu}{\sigma}\\Z=\frac{675-500}{100}$

Z=1.75

P(X>675)=1-P(X<675)=1-0.9599=0.0401

b)between 450 and 675?

P(450<X<675)

At x = 675

$Z=\frac{x-\mu}{\sigma}\\Z=\frac{675-500}{100}$

Z=1.75

At x = 450

$Z=\frac{x-\mu}{\sigma}\\Z=\frac{450-500}{100}$

Z=-0.5

P(450<X<675)=0.9599-0.3085=0.6514

Hence the probability that a randomly selected high school senior's score on mathematics part of SAT will be

(a) more than 675 is 0.0401

(b)between 450 and 675 is 0.6514

6 0

3 years ago

Guys please help me out. ...

Marina CMI [18]

I can't see the photo!????????

3 0

3 years ago

Marysya12 [62]

I think its option D

3 0

3 years ago

The verticals of a right triangle are (0, -6), (6, -3), and (x, -6). Find the value of x.

Nataly [62]

(0,-6), (6, -3), and (6,-6)

If you plot them on a graph there are only two point options (in green and purple) since it's a right triangle but you have to use the one that has -6 as the y value

6 0

3 years ago