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GrogVix [38]
4 years ago
8

B = what of the base

Mathematics
1 answer:
nevsk [136]4 years ago
7 0

Answer:

complete your question

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PLEASE HELP IT'S DUE TODAY
Korolek [52]
The initial value is 1
8 0
3 years ago
Situation:
stepan [7]

Answer:

t = 6162 years

Step-by-step explanation:

This equation takes the form of

N=N_{0}e^{kt}

We are given everything but the amount of C-14 at time t = 0.  But we can figure it out from the info we ARE given.  We are told that when the spear head is found it contains 54% of its original amount of C-14.  Notice we are dealing in percents.  If 54% remains when it is found, it started out with 100% of its amount.  That's our N value.  Filling in:

100=54e^{.0001t}

Our goal is to get that t down from the exponential position that it is currently in.  To do that we will need to eventually take the natural log of both sides, because ln's and e's "undo" each other, much like squaring "undoes" a square, or dividing "undoes" multiplication.  So we take the natural log of both sides.  On the right side, notice that when we take the natural log, the e disappears; it's "undone", gone.  Before that, though, we will simplify by dividing both sides by 54.  100/54 = 1.851851852.  So, altogether...

ln(1.851851852)=.0001t

Simplifying by plugging the log of that number into our calculator, we get

.6161861395 = .0001t  and

t = 6161.8613

Rounding to the nearest year, t = 6162 years

6 0
4 years ago
Please help!! I really don't want to get this one question wrong and mess up my score :(
Elan Coil [88]

Answer:

12^9 is the correct answer to the question

Step-by-step explanation:

12^8+(-)8-(-)9=8-8+9=9

12^9

7 0
3 years ago
Read 2 more answers
If the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb, how much work is needed to stretch it 6 in.
mel-nik [20]

Answer:

0.375 feet-lb

Step-by-step explanation:

We have been given that the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb. We are asked to find the work needed to stretch the spring 6 in. beyond its natural length.

We can represent our given information as:

6=\int\limits^2_0 {F(x)} \, dx

We will use Hooke's Law to solve our given problem.

F(x)=kx

Substituting this value in our integral, we will get:

6=\int\limits^2_0 {kx} \, dx

Using power rule, we will get:

6=\left[ \frac{kx^2}{2} \right ]^2_0

6=\frac{k(2)^2}{2}-\frac{k(0)^2}{2}

6=\frac{4k}{2}-0\\\\k=3

We know that 6 inches is equal to 0.5 feet.

Work needed to stretch it beyond 6 inches beyond its natural length would be \int\limits^{0.5}_0 {kx} \, dx =\int\limits^{0.5}_0 {3x} \, dx

Using power rule, we will get:

\int\limits^{0.5}_0 {3x} \, dx = \left [\frac{3x^2}{2}\right]^{0.5}_0

\frac{3(0.5)^2}{2}-\frac{3(0)^2}{2}\Rightarrow \frac{3(0.25)}{2}-0=\frac{0.75}{2}=0.375

Therefore, 0.375 feet-lb work is needed to stretch it 6 in. beyond its natural length.

3 0
3 years ago
Can someone help plz
Nata [24]

Answer:

the Horizontal Aysmptotes is: 4

Range is: (♾,4)

X intercept: None

Y intercept: (0,5)

Decreasing.

Step-by-step explanation:

Hope that will help you

4 0
3 years ago
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