An = a1 + (n-1)*d
n = term to find = 30
a1 = first term = 20
d = common difference = 2
a30 = 20 + (30 - 1) * 2
a30 = 20 + 29 * 2
a30 = 20 + 58
a30 = 78
sn = (n (a1 + a30)) / 2
s30 = (30(20 + 78) / 2
s30 = (30(98) / 2
s30 = 2940/2
s30 = 1470 <===
Answer:
b > 250/9 and w ≥ 0
Step-by-step explanation:
7/9 (b - 27) > 49/81
First, simplify both sides of the inequality:
7/9b - 21 > 49/81
Add 21 to both sides
7/9b - 21 +21 > 49/81 +21
Multiply both sides by 9/7
9/7 x 7/9b > 49/81 x 9/7
b > 1750/81
Simplify
b > 250/9
11w - 8w ≥ 14w
Simplify both sides once again
3w ≥ 14w
Subtract 14w from both sides
3w - 14w ≥ 14w - 14w
-11w ≥ 0
Divide both sides by -11w
w ≥ 0
I agree. The answer is choice B
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f ' (1) is shown to be positive
f ' (2) is shown to be positive as well
There are no critical values between x = 1 and x = 2, so f ' (x) is positive on the interval 1 < x < 2, so f(x) is increasing on this interval
All of this points to either choice A or choice B
--------------------------------
Similarly,
f ' (3) is negative
f ' (4) is negative
so f ' (x) is negative where 3 < x < 4 due to no other critical values being between x = 3 and x = 4
Based on these facts alone, the answer is either choice B or choice D
But we know that it can't be choice D as we determined it was between A and B. This rules out choice D
So all that's left is choice B
Answer:
6 and 7
Step-by-step explanation:
we know that 42 will lie between 36 and 49
And 36=6; 49=7
so we can be sure that 42 lies between 6 and 7
Answer:
Estimated mass of cooked food is 2267.96185 grams
Step-by-step explanation:
The total dinner cooked = 5 pounds
Now by the Parametric Conversion, we know that
1 pound = 453.59237 grams
So, converting the total mass of food in grams, we get
5 pound roast = 5 x 453.59237 grams roast
= 2267.96185 grams
⇒ 5 Pounds = 2267.96185 grams
So, estimated mass of cooked food is 2267.96185 grams.
