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inn [45]
3 years ago
9

Substatution and elimination

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
6 0
Well using elimination is easy because they basically did it for you. 
4x-5y=-6
2x+5y=12
The postive 5y and negative 5y cancel out because -5y plus 5y is zero.
so you are left with
4x=-6
2x=12
Add these together
6x=18
divide both sides by 6 and Solve to find your x
x=3
Now we have to eliminate x to find y.
we multiply the second equation by -2 so it can cancel out so
(2x+5y=12)-2
-4x-10y=24
Add the equations together
4x-5y=-6
-4x-10y=24
You are left with -15y = 30
Solve for y
y = -2
So your final answer for the elimination would be 
(3,-2)
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Answer:

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Step-by-step explanation:

1. 13=6x-2 (add 2 to each side)

2.15=6x (divide each side by 6)

3.15/6=x (simplify the fraction by dividing by 3)

4.5/2=x (answer)

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Triangle ABL is an isosceles triangle in circle A with a radius of 11, PL = 16, and ∠PAL = 93°. Find the area of the circle encl
katovenus [111]

Answer:

The area of the circle enclosed by line PL and arc PL is approximately 37.62 square units

Step-by-step explanation:

The given parameters in the question are;

The radius of the circle, r = 11

The length of the chord PL = 16

The measure of angle ∠PAL = 93°

The segment of the circle for which the area is required = Minor segment PL

The shaded area of the given circle is the minor segment of the circle enclosed by line PL and arc PL

The area of a segment of a circle is given by the following formula;

Area of segment = Area of the sector - Area of the triangle

In detail, we have;

Area of segment = Area of the sector of the circle that contains the segment) - (Area of the isosceles triangle in the sector)

Area of a sector = (θ/360)×π·r²

Where;

r = The radius of the circle

θ = The angle of the sector of the circle

Plugging in the the values of <em>r</em> and <em>θ</em>, we get;

The area of the sector enclosed by arc PL and radii AP and AL = (93°/360°) × π × 11² ≈ 98.2 square units

Area of a triangle = (1/2) × Base length × Height

Therefore;

The area of ΔAPL = (1/2) × 16 × 11 × cos(93°/2) ≈ 60.58 square units

∴ The area of the segment PL ≈ (98.2 - 60.58) square units = 37.62 square units

Therefore, the area of the shaded segment PL ≈ 37.62 square units

More examples on area of a shaded segment can be found here:

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