Answer:
Exercise 1:
base [b]=8cm
perpendicular [p]=6cm
hypotenuse [h]=?
<u>By</u><u> </u><u>using</u><u> </u><u>Pythagoras</u><u> </u><u>law</u>
h²=p²+b²
h²=6²+8²
h=√100
h=10cm
<u>So</u><u> </u><u>another</u><u> </u><u>side's</u><u> </u><u>length</u><u> </u><u>is</u><u> </u><u>1</u><u>0</u><u>c</u><u>m</u>
<u>Exercise</u><u> </u><u>2</u><u>:</u>
base [b]=6m
perpendicular [p]=bm
hypotenuse [h]=8m
By using Pythagoras law
h²=p²+b²
8²=b²+6²
b²=8²-6²
b=√28=2√7 0r 5.29 or 5.3
So height of kite is√<u>28</u><u>o</u><u>r</u><u> </u><u>2√7 0r 5.29 or 5.3 m</u>
Step-by-step explanation:
[Note: thanks for translating]
-18 because a negative number times positive is negative.
~JZ
C and D
Hope this helps !
Answer:
domain: x>3/5
Step-by-step explanation:
First we need to derive our function g(x) to get a new function g'(x)
To do this we will have to apply chain rule because we have an inner and outer functions.
Our G(x) = square root(3-5x)
Chain rule formula states that: d/dx(g(f(x)) = g'(f(x))f'(x)
where d/dx(g(f(x)) = g'(x)
g(x) is the outer function which is x^1/2
f(x) is our inner function which is 3-5x
therefore f'(x)= 1/2x^(-1/2) and f'(x) = -5
g'(f(x)) = -1/2(3-5x)^(-1/2)
Applying chain rule then g'(x) = 1/2 (3-5x)^(-/1/2)*(-5)
But the domain is the values of x where the function g'(x) is not defined
In this case it will be 3-5x > 0, because 3-5x is a denominator and anything divide by zero is infinity/undefined
which gives us x >3/5
