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kolbaska11 [484]
3 years ago
15

The area of a playground is 20 square yards. The length of the playground is 5 times longer than its width. Find the length and

width of the playground.
length = 1 yd, width = 20 yd
length = 2 yd, width = 10 yd
length = 10 yd, width = 2 yd
length = 20 yd, width = 1 y
Mathematics
1 answer:
jok3333 [9.3K]3 years ago
6 0
The answer is <span>length = 10 yd, width = 2 yd
proof 
</span><span>The length of the playground is 5 times longer than its width, it means
l=5w, but 10= 2x5 so we can write l= 5w, 
and Area = 20 =l x w = (5x2) x 2 =20</span>
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4 0
1 year ago
In JKL, k=4.1 cm,j=3.8 cm and angle J=Q3^ Find all possible values of angle K , to the nearest inch of a degree .
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Answer:

74.0°

Step-by-step explanation:

In triangle JKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°. Find all possible values of angle K, to the  nearest 10th of a degree

Solution:

A triangle is a polygon with three sides and three angles. Types of triangles are right angled triangle, scalene triangle, equilateral triangle and isosceles triangle.

Given a triangle with angles A, B, C and the corresponding sides opposite to the angles as a, b, c. Sine rule states that for the triangle, the following holds:

\frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}

In triangle JKL, k=4.1 cm, j=3.8 cm and angle J=63°.

Using sine rule, we can find ∠K:

\frac{k}{sin(K)}=\frac{j}{sin(J)}   \\\\\frac{4.1}{sin(K)}=\frac{3.8}{sin(63)}  \\\\sin(K)=\frac{4.1*sin(63)}{3.8}\\\\sin(K)=0.9613\\\\K=sin^{-1} (0.9613)\\\\K=74.0^o  \\

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