You have two 30-60-90 triangles, ADC and BDC.
The ratio of the lengths of the sides of a 30-60-90 triangle is
short leg : long leg : hypotenuse
1 : sqrt(3) : 2
Using triangle ADC, we can find length AC.
Using triangle BDC, we can find length BC.
Then AB = AC - BC
First, we find length AC.
Look at triangle ACD.
DC is the short leg opposite the 30-deg angle.
DC = 10sqrt(3)
AC = sqrt(3) * 10sqrt(3) = 3 * 10 = 30
Now, we find length BC.
Look at triangle BCD.
For triangle BCD, the long leg is DC and the short leg is BC.
BC = 10sqrt(3)/sqrt(3) = 10
AB = AC - BC = 30 - 10 = 20
There are different definitions of "whole numbers".
Some define it as an integer (i.e. positive or negative) [some dictionaries]
Some define it as a non-negative integer. [most math definitions]
We will take the math definition, i.e. 0<= whole number < ∞
To find pairs (i.e. two) whole numbers with a sum of 110, we start with
0+110=110
1+109=110
2+108=110
...
54+56=110
55+55=110
Since the next one, 56+54=110 is the same pair (54,56) as 54+56=110, we stop at 55+55=110 for a total of 56 pairs.
Answer: d.h=−4
PLZ MARK BRAINLIEST!
Step-by-step explanation:
Let's solve your equation step-by-step.
−3(h+5)+2=4(h+6)−9
Step 1: Simplify both sides of the equation.
−3(h+5)+2=4(h+6)−9
(−3)(h)+(−3)(5)+2=(4)(h)+(4)(6)+−9(Distribute)
−3h+−15+2=4h+24+−9
(−3h)+(−15+2)=(4h)+(24+−9)(Combine Like Terms)
−3h+−13=4h+15
−3h−13=4h+15
Step 2: Subtract 4h from both sides.
−3h−13−4h=4h+15−4h
−7h−13=15
Step 3: Add 13 to both sides.
−7h−13+13=15+13
−7h=28
Step 4: Divide both sides by -7.
−7h
−7
=
28
−7
h=−4
Step-by-step explanation:
Since f(0) = f(5) = f(8) = 0, we have f(x) = Ax(x - 5)(x - 8), where A is a real constant.
We know that f(10) = 17.
=> A(10)(10 - 5)(10 - 8) = 17
=> A(10)(5)(2) = 17
=> 100A = 17, A = 0.17.
Hence the answer is f(x) = 0.17x(x - 5)(x - 8).
We know that
Half-life is modeled by the formula
An=A0*(0.5)<span>^[t/h)]
where
An----------> </span>is the amount remaining after a time t
A0----------> is the initial quantity
t------------> is the time
h------------> is the half-life of the decaying quantity
in this problem
h=1601 years
A0=50 g
An=?
t=100 years
An=A0*(0.5)^[t/h)]---------> An=50*(0.5)^[100/1601)]-----> 47.88 gr
the answer is 47.88 g
