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Alex
3 years ago
12

X^2=4 square root property

Mathematics
1 answer:
Olin [163]3 years ago
7 0

{x}^{2}  = 4 \\  x  =  \sqrt{4}  \\  x = 2
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1. Find the cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of ₹ 3 per m
Flauer [41]

1. Find the cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of ₹ 3 per m².

a = 51 m, b = 37 m, c = 20 m

semiperimeter:  p = (51+37+20):2 = 54 m

Area of triangle:

A=\sqrt{p(p-a)(p-b)(p-c)}\\\\A=\sqrt{54(54-51)(54-37)(54-20)}\\\\A=\sqrt{54\cdot3\cdot17\cdot34}\\\\A=\sqrt{9\cdot2\cdot3\cdot3\cdot17\cdot17\cdot2}\\\\A=3\cdot2\cdot3\cdot17\\\\A=306\,m^2

Rate:  ₹ 3 per m².

Cost:  ₹ 3•306 = ₹ 918

2. Find the area of the isosceles triangle whose perimeter is 11 cm and the base is 5 cm.

a = 5 cm

a+2b = 11 cm  ⇒  2b = 6 cm   ⇒  b = 3 cm

p = 11:2 = 5.5

A=\sqrt{5.5(5.5-3)^2(5.5-5)}\\\\ A=\sqrt{5.5\cdot(2.5)^2\cdot0.5}\\\\ A=\sqrt{11\cdot0.5\cdot(2.5)^2\cdot0.5}\\\\A=0.5\cdot2.5\cdot\sqrt{11}\\\\A=1.25\sqrt{11}\,cm^2\approx4.146\,cm^2

3. Find the area of the equilateral triangle whose each side is 8 cm.

a = b = c = 8 cm

p = (8•3):2 = 12 cm

A=\sqrt{12(12-8)^3}\\\\ A=\sqrt{12\cdot4^3}\\\\ A=\sqrt{3\cdot4\cdot4\cdot4^2}\\\\A=4\cdot4\cdot\sqrt{3}\\\\A=16\sqrt3\ cm^2\approx27.713\ cm^2

4. The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle.

a = 2x

b = 3x

2x + 2•3x = 32 cm  ⇒ 8x = 32 cm   ⇒  x = 4 cm  ⇒ a = 8 cm, b = 12 cm

p = 32:2 = 16 cm

A=\sqrt{16(16-8)(16-12)^2}\\\\ A=\sqrt{16\cdot8\cdot4^2}\\\\ A=\sqrt{2\cdot8\cdot8\cdot4^2}\\\\ A=8\cdot4\cdot\sqrt2\\\\ A=32\sqrt2\ cm^2\approx45.2548\ cm^2

8 0
3 years ago
Please help me if u can!!​
damaskus [11]

Answer:

89 kids are in the 3 rd grade

Step-by-step explanation:

3 0
4 years ago
Read 2 more answers
George is given two circles, circle O and circle X 1 as shown. If he wants to prove that the two circles are similar, what would
salantis [7]

Answer:

Step-by-step explanation:

Given: The radius of circle O is r, and the radius of circle X is r'.

To prove: Circle O is similar to circle X.

Proof: Move the center of the smaller circle onto the center of the largest circle. Translate the circle X by the vector XA onto circle O. The circles now have the same center.

A dilation is needed to increase the size of circle X to coincide with the circle O. A value which when multiplied by r' will create r.

The scale factor x to increase X:

⇒

A translation followed by a dilation with scale factor  will map one circle to the other, thus proving the given both circles similar.

Therefore, circle O is similar to circle X.

Step-by-step explanation:

6 0
3 years ago
3(5b-1)=5<br> solve for b<br><br> 4(3y-1)=20<br> solve for y
Gnom [1K]
3(5b-1)=5
Get rid of the brackets
15b-3=5
add 3 to both sides of the equation
15b=8
divide both sides of the equation by 15
b=0.53

4(3y-1)=20
12y-4=20
12y=24
y=0.5
7 0
3 years ago
HELP NEEDED ASAP MATHS PLEASE!!!
BabaBlast [244]
You plug in the values for x in each respective equation.

a) f(3) = 3 + 1 which equals: 4.
so, f(3) = 4

b) g(3) = (3)^2 - 3 which equals 9 - 3, which equals 6.
so, g(3) = 6.

hope this helps! (:
5 0
3 years ago
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