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nataly862011 [7]
3 years ago
11

Y=x-4 2x+y=5 can you please help me with this math problem?????

Mathematics
1 answer:
OleMash [197]3 years ago
7 0
X=1
y=-3
i used a graph to work this out.
hope this helps. don't hesitate to tell me if it doesn't :)

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Sav [38]

The area of the circle is 28 ft square.

The area of the triangle is 30 ft square.

Step-by-step explanation:

1) The area of the circle= π*r^2

where π= 3.14(default value) and "r" is the radius of the circle.

Here, Radius of the circle(r)= 3 ft

Area = 3.14*3*3

        = 28.26 ft sq.= 28 ft sq(rounded to the nearest whole number)

2) The triangle has the base= 10 feet and height= 6 feet

The Area of the triangle= 1/2(b)(h)

where "b" is the base and "h" is the height of the triangle.

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4 years ago
How can you get equation b from equation A. 3(x+2)=18 B. X+2=18
Rzqust [24]

Answer:

We start with the equation:

A: 3*(x + 2) = 18

And we want to construct equation B:

B: X + 2 = 18

where I suppose that X is different than x.

Because in both equations the right side is the same thing, then the left side also should be the same thing, this means that:

3*(x + 2) = X + 2

Now we can isolate the variable x.

(x + 2) = (X + 2)/3

x = (X + 2)/3 - 2

Then we need to replace x by  (X + 2)/3 - 2 in equation A, and we will get equation B.

Let's do it:

A: 3*(x + 2) = 18

Now we can replace x by =  (X + 2)/3 - 2

3*( (X + 2)/3 - 2 + 2) = 18

3*(  (X + 2)/3 ) = 18

3*(X + 2)/3 = 18

(X + 2) = 18

Which is equation B.

3 0
3 years ago
Simplify using trigonometric identities<br> 2sinθ - sin2θ cosθ
Anit [1.1K]

Double angle identity for sine:

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Factorize the left side.

2 \sin(\theta) - 2 \sin(\theta) \cos^2(\theta) = 2 \sin(\theta) \left(1 - \cos^2(\theta)\right)

Pythagorean identity:

\cos^2(x) + \sin^2(x) = 1

\implies 2 \sin(\theta) \left(1 - \cos^2(\theta)\right) = 2 \sin^3(\theta)

so that

\boxed{2 \sin(\theta) - \sin(2\theta) \cos(\theta) = 2 \sin^3(\theta)}

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Divide the long leg by the square root of 3 to find the short leg. Double that figure to find the hypotenuse.
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