To calculate for the area of the triangle given all its side, use the Heron's formula which is,
A = sqrt ((s)(s - a)(s - b)(s - c))
where A is the area. a, b, and c are the measure of the sides and s the semi - perimeter or half of the perimeter.
With the given above, s = (17 + 18 + 21.4) / 2 = 28.2
Substituting the known dimensions to the Heron's formula,
A = sqrt ((28.2)(28.2 - 17)(28.2 - 18)(28.2 - 21.4)) = 148 units^3
Thus, the area of the triangle is 148 units^3.
160 = 2 x 2 x 2 x 2 x 2 x 5 = 2^5 x 5
243 = 3 x 3 x 3 x 3 x 3 = 3^5
So
(160 * 243)^1/5
= 5th root of (160 * 243)
= 5th root of (2^5 * 5 * 3^5)
= 2 * 3 * (5th root of 5)
= 6 * (5th root of 5)
Answer is D. 6 * (5th root of 5)
7436 ÷ 13 = 572
I didn't understand the rest of the thing u mentioned
Answer:
- a° = 70°
- b° = 78°
- c° = 139°
- d° = 17°
- e° = 123°
- f° = 69.5°
Step-by-step explanation:
You need to take advantage of what you have learned about inscribed angles. Each inscribe angle is half the measure of the arc it intercepts. Of course, the sum of all arcs around a circle is 360°.
Putting these facts together, you can quickly conclude that opposite angles of an inscribed quadrilateral are supplementary:
a = 180 -110 = 70
b = 180 -102 = 78
The arc (c+81) is opposite the inscribed angle marked 110, so we have ...
c + 81 = 2(110)
c = 220 -81 = 139
The arc (c+d) is intercepted by the angle marked b, so we have ...
139 +d = 2(78)
d = 156 -139 = 17
The arc (e+81) is intercepted by the inscribed angle marked 102, so ...
e +81 = 2(102)
e = 204 -81 = 123
The inscribed angle f intercepts arc c, so is half its measure.
f = 139/2 = 69.5