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sergiy2304 [10]
3 years ago
15

HELP. SUPER EASY MATH PROBLEM. 20 POINTS

Mathematics
2 answers:
Stells [14]3 years ago
7 0

The area of the shaded = the area of the sector - the area of the isosceles triangle.

Split the isosceles triangle in half, each of those triangles is a 30-60-90 triangle, where the radius is the hypotenuse.

Ratio of short leg: long leg: hypotenuse= x : x√3 : 2x

Given hypotenuse = radius = 27.8 in

so short leg = 27.8 / 2 = 13.9 in

long leg = 13.9 √3 = 24.1 in

Area of this isosceles triangle = 24.1 x 13.9 = 334.99 in^2

Area of sector = (150 / 360) π r^2

= (150 / 360) (3.14) (27.8)^2

= 0.4 * (3.14) (772.84)

= 970.7 in^2

Area of shaded = 970.7 in^2 - 334.99 in^2 = 635.71 in^2

= 635.7 in^2 (nearest tenth)

Answer:

635.7 in^2

Stells [14]3 years ago
6 0

First find the area of the sector of 150 degrees:

150/360 x PI * 27.8^2 = 1011.6452 in^2

Now find the area of the triangle formed by the 150 degrees and the radius 27.8

first find the length of the chord ( the base of the triangle)

c^2 = 27.8^2 + 27.8^2 - 2 * 27.8 * cos(150) = 53.71

now find the height by using half the chord length as the base and using the Pythagorean theorem:

26.855^2 - 27.8^2 = height^2

height = 7.19

The area of the triangle is 1/2 x base x height = 1/2 x 26.855 x 7.1952 = 96.605

multiply by 2 because we made 2 right triangles: 96.54 x 2 = 193.21 square inches.

Now the area of the shaded region is the area of the sector minus the area of the triangle:

1011.6452 - 193.21 = 818.4352

Rounded to nearest tenth = 818.4 in^2

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If JKLM is a rhombus, MK = 30, NL = 13, and mZMKL = 41°, find each measure.
oksian1 [2.3K]

Answer:

NK = 15

JL = 26

KL = 19.85

\angle JKM =49

\angle JML =41

\angle MLK = 90

\angle MNL =90

\angle KJL =41

Step-by-step explanation:

Given

MK = 30

NL = 13

\angle MKL = 41

Solving (a): NK

MK is a diagonal and NK is half of the diagonal. So:

NK = \frac{1}{2} * MK

NK = \frac{1}{2} * 30

NK = 15

Solving (b): JL

JL is a diagonal, and it is twice of NL.

JL = 2 * NL

JL = 2 * 13

JL = 26

Solving (c): KL

To solve for KL, we consider triangle KNL where:

\angle KNL = 90

and

KL^2 = NL^2 + NK^2

KL^2 = 13^2 + 15^2

KL^2 = 394

KL = \sqrt{394

KL = 19.85

Solving (d - h):

To do this, we consider triangle JKN

\angle KNL = \angle LNM = \angle MNJ = \angle JNK = 90 -- diagonals bisect one another at right angle

Alternate interior angles are equal. So:

\angle MKL = \angle KMJ = \angle KJL = \angle JLM = 41

Similarly:

\angle MKJ = \angle KML = \angle MJL = \angle JLK = 90 - 41

\angle MKJ = \angle KML = \angle MJL = \angle JLK = 49

So:

\angle JKM =49

\angle JML =41

\angle MLK = \angle MLJ + \angle JLK

\angle MLK = 49 + 41

\angle MLK = 90

\angle MNL =90

\angle KJL =41

5 0
3 years ago
Recall that the volume of a sphere is given by the formula V = 4/3pi to the 3rd
Kisachek [45]

9514 1404 393

Answer:

  1. (256/3)π ft³

  2. 33.49 ft³

Step-by-step explanation:

1. Put 4 ft into the formula where r is, then simplify.

  V = (4/3)π(4 ft)³ = (256/3)π ft³

__

2. The radius is half the diameter, so is 2 ft. Put that in the formula, along with the value for π, and do the arithmetic.

  V = (4/3)(3.14)(2 ft)³ ≈ 33.49 ft³

4 0
3 years ago
The area of a trapezium is 360 cm² and its height is 18 cm. If one of the parallel sides is longer than the other by 6 cm, then
Leokris [45]

Answer:

The two parallel sides a and b are 17 cm and 23 cm respectively

Step-by-step explanation:

Area of a trapezium = {(a+b)/2} h

Where

h = height = 18 cm

a = parallel side 1 = x

b = parallel side 2 = x + 6

Area of a trapezium = {(a+b)/2} h

360 = {(x + x + 6) / 2} 18

360 = { (2x + 6) / 2} 18

360 = (2x + 6) 9

360 = 18x + 54

Subtract 54 from both sides

360 - 54 = 18x

306 = 18x

Divide both sides by 18

x = 306 / 18

= 17

x = 17 cm

a = parallel side 1 = x

x = 17 cm

b = parallel side 2 = x + 6

x + 6

= 17 + 6

= 23 cm

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3 years ago
What is the y-intercept of the line represented by the equation y = 2x - 3?
cestrela7 [59]
Y =Mx + C 
where M = gradient / slope
           C = y intercept

comparing with y = 2x - 3
M = 2
C = -3 
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3 years ago
Which is a good comparison of the estimated sum and the actual sum of 7 7/8 + 2 2/11?
Ipatiy [6.2K]
7 7/8 can be estimated to 8

2 2/11 can be estimated to 2

8 + 2 = 10


7 7/8 = 56+7/8 = 63/8

2 2/11 = 22+2/11 = 24/11

63/8 + 24/11 = 693+192/88 = 10.06

So do you see how similar 10 and 10.06 are to each other? :)

3 0
3 years ago
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