Answer:
70. 0
71. -54
72. 12
73. 86
74. 59
Step-by-step explanation:
To evaluate an expression, substitute specific values for the variables and simplify using Order of Operations.
70. c-3d becomes
71. 
becomes
72. 
becomes
73. 
becomes
74. becomes
We will use demonstration of recurrences<span>1) for n=1, 10= 5*1(1+1)=5*2=10, it is just
2) assume that the equation </span>10 + 30 + 60 + ... + 10n = 5n(n + 1) is true, <span>for all positive integers n>=1
</span>3) let's show that the equation<span> is also true for n+1, n>=1
</span><span>10 + 30 + 60 + ... + 10(n+1) = 5(n+1)(n + 2)
</span>let be N=n+1, N is integer because of n+1, so we have
<span>10 + 30 + 60 + ... + 10N = 5N(N+1), it is true according 2)
</span>so the equation<span> is also true for n+1,
</span>finally, 10 + 30 + 60 + ... + 10n = 5n(n + 1) is always true for all positive integers n.
<span>
</span>
Answer:
the answer is s=5/4 minus 1/16
hope this helps :D
Check the picture below.
since the focus point is above the directrix, that simply means the parabola is a vertical one, and therefore the square variable is the "x".
keeping in mind that, there's a distance "p" from the vertex to either the focus point or the directrix, that puts the vertex half-way between those fellows, in this case at 0, right between 1 and -1, as you see in the picture, and the parabola looks like so.
since the parabola is opening upwards, the value for "p" is positive, thus
3.2 x 108 The 8 is suppose to be an exponent of 10
