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Maru [420]
3 years ago
11

n ΔXYZ, m∠X = 90° and m∠Y = 30°. In ΔTUV, m∠U = 30° and m∠V = 60°. Which is true about the two triangles? ΔXYZ ≅ ΔTUV ΔXYZ ≅ ΔVU

T No congruency statement can be made because only two angles in each triangle are known. No congruency statement can be made because the side lengths are unknown.
Mathematics
2 answers:
Nikolay [14]3 years ago
6 0

Answer:

<em><u>D. No congruency statement can be made because the side lengths are unknown.</u></em>

Step-by-step explanation:

I JUST TOOK THE TEST!!!!

lions [1.4K]3 years ago
5 0
The triangles are similar because corresponding angles are congruent, however no congruency statement can be made because the side lengths are unknown.

The 4th selection is appropriate.
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The price of a toy usually costing £50 is increases to £65<br> work out the percentage increase
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Answer:

30%

Step-by-step explanation:

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3 years ago
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3 years ago
In Exercise 4, find the surface area of the solid<br> formed by the net.
Fittoniya [83]

Answer:

3. 150.72 in²

4. 535.2cm²

Step-by-step Explanation:

3. The solid formed by the net given in problem 3 is the net of a cylinder.

The cylinder bases are the 2 circles, while the curved surface of the cylinder is the rectangle.

The surface area = Area of the 2 circles + area of the rectangle

Take π as 3.14

radius of circle = ½ of 4 = 2 in

Area of the 2 circles = 2(πr²) = 2*3.14*2²

Area of the 2 circles = 25.12 in²

Area of the rectangle = L*W

width is given as 10 in.

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Area of rectangle = L*W = 12.56*10 = 125.6 in²

Surface area of net = Area of the 2 circles + area of the rectangle

= 25.12 + 125.6 = 150.72 in²

4. Surface area of the net (S.A) = 2(area of triangle) + 3(area of rectangle)

= 2(0.5*b*h) + 3(l*w)

Where,

b = 8 cm

h = \sqrt{8^2 - 4^2} = \sqrt{48} = 6.9 cm} (Pythagorean theorem)w = 8 cm[tex]S.A =  2(0.5*8*6.9) + 3(20*8)

S.A =  2(27.6) + 3(160)

S.A =  55.2 + 480

S.A = 535.2 cm^2

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3 years ago
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Can you help me with this question please
solniwko [45]

Answer:

the answer is 9^8

Step-by-step explanation:

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3 years ago
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