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const2013 [10]
3 years ago
15

Please answer these you don’t need to show your work just answers please!!

Mathematics
1 answer:
boyakko [2]3 years ago
5 0
#9- -1.42
#10- .04
#11- .232
#12- 6.6
#13- 4 505/1000
#14- -11/100
#15- 62/100
#16- -1 74/100
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Craig’s June water bill listed two meter reading. The previous reading was 6,372 ccf and the present reading is 6501 CFC. How mu
romanna [79]

Answer: Craig's household use of water during the billing period = 129 CFC

Explanation:

Since we have given that

Craig's June water meter's previous reading =6372 CFC

Craig's water meter's present reading =6501 CFC

Craig's household use during the billing period =6501 -6372= 129 CFC

So, we get that

Craig's household use of water during the billing period = 129 CFC


8 0
3 years ago
Who got the answers for lib arts 1 flvs 1.07??
aleksklad [387]

Answer:

is the answer is 1.07 to 7.0

Step-by-step explanation:


8 0
3 years ago
This is from gradpoint, please help me y’all :((
Afina-wow [57]

Answer if im not misstaken its no

Step-by-step explanation:

because ur working with fractions they dont = the same thing. lmk if im wrong

5 0
2 years ago
Read 2 more answers
For fun question
ZanzabumX [31]
Consider the function f(x)=x^{1/3}, which has derivative f'(x)=\dfrac13x^{-2/3}.

The linear approximation of f(x) for some value x within a neighborhood of x=c is given by

f(x)\approx f'(c)(x-c)+f(c)

Let c=64. Then (63.97)^{1/3} can be estimated to be

f(63.97)\approxf'(64)(63.97-64)+f(64)
\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375

Since f'(x)>0 for x>0, it follows that f(x) must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function f(x). This means the estimated value is an overestimation.

Indeed, the actual value is closer to the number 3.999374902...
4 0
3 years ago
Christine bakes 4 dozen cookies in an hour. How many dozen cookies does she bake in 3.5 hours?
Paladinen [302]

Answer:

14 dozen

Step-by-step explanation:

multiply 4x3.5=14

7 0
3 years ago
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