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galina1969 [7]
3 years ago
15

Use the distributive property to rewrite each algebraic expression.

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
3 0

Answer:

2

Step-by-step explanation:

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In triangle JKL, sin(b°) = three fifths and cos(b°) = four fifths. If triangle JKL is dilated by a scale factor of 2, what is ta
scoundrel [369]

Answer:

Tan(b°) = 3/4 = three fourths (C)

Step-by-step explanation:

Triangle JKL is a right angled triangle in which angle K is a right angle and angle L= b°

We would apply SOHCAHTOA from trigonometry to determine the value for each sides.

For ∆JKL

Sin(b°) = opposite/hypotenuse

Sin(b°) = 3/5

Cos(b°) = adjacent/hypotenuse

Cos(b°) = 4/5

Tan(b°) = 3/4

From the above we know the value of each sides of ∆JKL

Find attached the diagram of the above triangle (1)

∆JKL is dilated by a scale factor = 2

Meaning we would multiply each sides of ∆JKL by 2. The values of the new triangle become:

Opposite = 2(3) = 6

Adjacent = 2(4) = 8

Hypotenuse = 2(5) = 10

To find tan(b°) of the new triangle, we would apply tangent ratio

Tan(b°) = opposite/adjacent

Tan(b°) = 6/8

Tan(b°) = 3/4

Find attached the diagram for the new triangle (2)

Diagram 3 shows the drawing of both triangles together.

From the above, we can see the angle doesn't change when a shape is dilated by a scale factor.

Therefore, tan(b°) = three fourths (C)

6 0
3 years ago
Area of regular polygons
Finger [1]
Hello!

To find the area of a hexagon you do 3\frac{3 \sqrt{3} }{2} } a^{2} where a is one of the sides

Since the perimeter is 60 we can do 60/ the sides of the shape

60/6 = 10

So one side is equal to 10

you put that into the formula to get the area of a hexagon and you get 259.81

Hope this helps!
8 0
3 years ago
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