Check the picture below.
notice the sides, now, on the second triangle, side 6 slants a bit more to fit in 13, on the third triangle, side 6 slants even further to fit 13 in, now, if 6 were to slant completely, it'll make a flat-line with side 5, and there will be a triangle no more.
but even if side 6 would stretch to a flat-line, 5+6 is just 11, whilst side 13 is longer than that, so no dice.
Answer:
The answer is a. 21.
Step-by-step explanation:
We solve this question by using Pythagoras theorem to relate the sides of a right angled triangle.
Here,
Hypotenuse of the given triangle(c)=75
Two sides of right angled triangle are:
a=72
b=?
Then,
for a given right angled triangle abc,
Using Pythagoras theorem,
![c=\sqrt{a^{2} +b^{2} }](https://tex.z-dn.net/?f=c%3D%5Csqrt%7Ba%5E%7B2%7D%20%2Bb%5E%7B2%7D%20%7D)
Squaring on both sides,
![c^{2} =a^{2}+b^{2}](https://tex.z-dn.net/?f=c%5E%7B2%7D%20%3Da%5E%7B2%7D%2Bb%5E%7B2%7D)
or,![75^{2}=72^{2}+b^{2}](https://tex.z-dn.net/?f=75%5E%7B2%7D%3D72%5E%7B2%7D%2Bb%5E%7B2%7D)
or, ![b^{2}=75^{2}-72^{2}](https://tex.z-dn.net/?f=b%5E%7B2%7D%3D75%5E%7B2%7D-72%5E%7B2%7D)
or,![b^{2}=5625- 5184](https://tex.z-dn.net/?f=b%5E%7B2%7D%3D5625-%205184)
or,![b^{2} =441](https://tex.z-dn.net/?f=b%5E%7B2%7D%20%3D441)
∴![b=21](https://tex.z-dn.net/?f=b%3D21)
So, the value of b is obtained to 21 by the use of Pythagoras theorem.
Step-by-step explanation:
the circumference of a circle is
C = 2×pi×r
with r being the radius.
the square side length is 22 cm. so, the circumference or perimeter of the square (and therefore the length of the wire) is
4×22 = 88 cm.
now, it is the same wire of the same length that is now forming a circle.
so, the circumference of the square is also the circumference of the circle.
therefore,
88 = 2×pi×r
44 = pi×r
r = 44/pi = 14.00563499... cm
so, rounded, radius = 14 cm.
MARK BRAINLIEST!!
So your solution is.
=−5
Answer:
(4×9+5) -10+16×6
36 + 5 - 10 + 16 * 6
41 - 10 + 96
41 + 86
127
Step-by-step explanation:
(4×9+5) -10+16×6
36 + 5 - 10 + 16 * 6
41 - 10 + 96
41 + 86
127
use PEMDAS
p = parantheses
e= exponent
m = multiply
d = divide
a = add
s = subtract