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lakkis [162]
3 years ago
11

Which triangle is it ?

Mathematics
1 answer:
coldgirl [10]3 years ago
5 0
The answer is: The first triangle. The reasons are shown below:

 1. All the triangles are rigth triangles, because they have an angle of 90°. So, let's calculate the others angles of the first one:

 Tan(α)^-1= opposite leg/adjacent leg

 Opposite leg=5
 Adjacent leg=5√3

 Tan(α)^-1= 5/5√3
 Tan(α)^-1=30°

 2. Let's calculate the other angle:

 Tan(α)^-1= opposite leg/adjacent leg

 Now, the opposite leg will be 5√3 and the adjacent leg will be 5. Then:

 Tan(α)^-1= 5√3/5
 Tan(α)^-1=60°

 As you can see, the angles of first triangle are: 30°,60° and 90°. 
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It is C. floor 20 and floor c have the same amount of tenants
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Find the length of arc ac in terms of pi
aniked [119]
=theta/360°×2pi ×r
theta=180-60=120°
r=24÷2=12

120/360 × 2 × 22/7 × 12=
1/3 × 2 × 22/7 × 12=
176/7=
25.14


3 0
4 years ago
Find an expression (where you have to multiply polynomials) to represent the shaded region of #33 and if you could also include
Brrunno [24]
First, find the area of the circle using the formula A=pi r^2. 
A=pi (2x+3)^2 = pi(4x^2 + 12x + 9) 
Second, find the area of the rectangle inside by multiplying the polynomials. 
(X+1)*(3x+2) = 3x^2 + 5x +2 
Third, subtract the area of the rectangle from the area of the circle to find the area of the shaded region. 
pi(4x^2 + 12x + 9) - 3x^2 + 5x +2 =area of shaded region 
Or 
(pi (2x+3)^2) - ((X+1)(3x+2)) = area of shaded region


4 0
3 years ago
Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first
BaLLatris [955]

Answer:

\cos(x+y) goes with -\frac{\sqrt{6}+\sqrt{2}}{4}

\sin(x+y) goes with \frac{\sqrt{6}-\sqrt{2}}{4}

\tan(x+y) goes with \sqrt{3}-2

Step-by-step explanation:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

We are given:

\sin(x)=\frac{\sqrt{2}}{2} which if we look at the unit circle we should see

\cos(x)=\frac{\sqrt{2}}{2}.

We are also given:

\cos(y)=\frac{-1}{2} which if we look the unit circle we should see

\sin(y)=\frac{\sqrt{3}}{2}.

Apply both of these given to:

\cos(x+y)

\cos(x)\cos(y)-\sin(x)\sin(y) by the addition identity for cosine.

\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}

\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}

\frac{-\sqrt{2}-\sqrt{6}}{4}

-\frac{\sqrt{6}+\sqrt{2}}{4}

Apply both of the givens to:

\sin(x+y)

\sin(x)\cos(y)+\sin(y)\cos(x) by addition identity for sine.

\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}

\frac{-\sqrt{2}+\sqrt{6}}{4}

\frac{\sqrt{6}-\sqrt{2}}{4}

Now I'm going to apply what 2 things we got previously to:

\tan(x+y)

\frac{\sin(x+y)}{\cos(x+y)} by quotient identity for tangent

\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}

-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}

-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}

-\frac{8-2\sqrt{12}}{4}

There is a perfect square in 12, 4.

-\frac{8-2\sqrt{4}\sqrt{3}}{4}

-\frac{8-2(2)\sqrt{3}}{4}

-\frac{8-4\sqrt{3}}{4}

Divide top and bottom by 4 to reduce fraction:

-\frac{2-\sqrt{3}}{1}

-(2-\sqrt{3})

Distribute:

\sqrt{3}-2

6 0
3 years ago
I will give brainliest if anyone knows this
mezya [45]

Answer:

7 4

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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