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Anvisha [2.4K]
3 years ago
14

Please help it will be brainlist please make sure you are 100% the awnser is right

Mathematics
1 answer:
Dominik [7]3 years ago
4 0

Answer:

Step-by-step explanation:

Step 1: Sum of angles on a straight line is 180

Step 2:

2x + 25 + y = 180

2x + y = 180 - 25

2x + y = 155  (1)

Step 3:

3x - 10 + y = 180

3x + y = 180 + 10

3x + y = 190    (2)

Step 4: Substract equation 1 from 2

3x + y - 2x - y = 190 - 155

      x = 35

Step 5:

Substitute x in equation 1 to find y

2x + y = 55

2(35) + y = 155

70  +  y = 155

  y = 155 - 70

   y = 85

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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
What is 2/3 divide by 2/9
fenix001 [56]

Answer:

3

Step-by-step explanation:

(2/3)/(2/9) = (2/3) * (9/2) = 3

8 0
3 years ago
Micheal has 5x as much money minus 25 as mia. togther they have 79.
klemol [59]

Answer:

10

Step-by-step explanation:

5 10 / 15 20 15.......

4 0
3 years ago
A car traveling at constant speed traveled 262.5 miles in 3 1⁄2 hours. What was the car's speed?
CaHeK987 [17]
S= 262.5÷ 3.5 = 75 mph
5 0
3 years ago
Read 2 more answers
HELP PLS GIVING BRAINLIEST
Setler79 [48]

For this case we have by definition, that the equation of a line in the slope-intersection form is given by:

y = mx+b

Where:

m: It's the slope

b: It is the cutoff point with the y axis

We need two points through which the line passes to find the slope:

(0,1)\\(1, -2)

We found the slope:

m = \frac {y2-y1} {x2-x1}\\m = \frac {-2-1} {1-0} = \frac {-3} {1} = - 3

So, the equation is of the form:

y = -3x + b

We substitute a point to find "b":

1 = -3 (0) + b\\1 = b

Finally, the equation is:

y = -3x+1

Answer:

Option C

8 0
3 years ago
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