1. The area of the full rectangle is 21 * 28 = 588. We will need this later.
Like you drew, there is an imaginary triangle missing. We can find the dimensions of it by subtracting from the given information. The shorter leg of it is 21 - 12 = 9, the longer leg is 28 - 14 = 14.
Now we find the area of the triangle. Area of a triangle is 1/2 * base * height. So 1/2 * 9 * 14 = 1/2 * 126 = 63.
Now we do the full area of 588 minus the area of the missing triangle piece.
588 - 63 = 525 ft²
2. Do the area of the rectangle and then this time add on the area of the triangle. 7 * 15 = 105. Area of the triangle = 1/2 * 2 * 7 = 1/2 * 14 = 7.
105 + 7 = 112 in²
Answer:
a
Step-by-step explanation:
<span>A dust particle weighs 7.42 × 10-10 kilograms.
Weight of 5 × 106 dust particles = </span>5 × 10^6 x 7.42 x 10^(-10)
= 5 x 7.42 x 10^(6-10)
= 37.1 x 10^(-4)
= 3.71 x 10^(-3) kilograms
Answer:
Step-by-step explanation:
The formula of a volume of a sphere:
We have
Substitute:
<em>divide both sides by π</em>
<em>multiply both sides by 3</em>
<em>divide both sides by 4</em>
![R^3=\dfrac{1}{2}:4\\\\R^3=\dfrac{1}{2}\cdot\dfrac{1}{4}\\\\R^3=\dfrac{1}{8}\to R=\sqrt[3]{\dfrac{1}{8}}\\\\R=\dfrac{\sqrt1}{\sqrt8}\\\\R=\dfrac{1}{2}](https://tex.z-dn.net/?f=R%5E3%3D%5Cdfrac%7B1%7D%7B2%7D%3A4%5C%5C%5C%5CR%5E3%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%5Cdfrac%7B1%7D%7B4%7D%5C%5C%5C%5CR%5E3%3D%5Cdfrac%7B1%7D%7B8%7D%5Cto%20R%3D%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B8%7D%7D%5C%5C%5C%5CR%3D%5Cdfrac%7B%5Csqrt1%7D%7B%5Csqrt8%7D%5C%5C%5C%5CR%3D%5Cdfrac%7B1%7D%7B2%7D)
Answer:
<u><em>Question 10.</em></u>
Length of the side of the square = 
By Pythagoras' Theorem,
Hypotenuse = 
(H > 0)
= 29 m
Ans: 29 m
<u><em>Question 11.</em></u>
By Pythagoras' Theorem,
Length of longer side of 42 km = 42- 
(L > 0)
= 32.050 km (5 s.f.)
By Pythagoras' Theorem,
Length of the unknown side = 
(L > 0)
= 35.4 km (3 s.f.)
Ans: 35 km
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