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GuDViN [60]
3 years ago
10

Help me figure this out please it’s due today i will name you brainliest!

Mathematics
1 answer:
kompoz [17]3 years ago
7 0

Answer:

i know 4 is 1 i think

Step-by-step explanation:

You might be interested in
Which is an exponential decay function f (x)=3/4 (7/4)x f (x)=2/3 (4/5)-x. F (x)=3/2 (8/7)-x f (x)=1/3 (9/2)x
dsp73
The given functions are
1.\, f(x)= \frac{3}{4}( \frac{7}{4})^{x} \\ \\2.\,f(x)= \frac{2}{3}( \frac{4}{5} ) ^{-x}\\\\3.\, f(x)= \frac{3}{2} ( \frac{8}{7} )^{-x}\\\\4.\,f(x)= \frac{1}{3} ( \frac{9}{2} )^{x}

Evaluate the functions. 
1. Because 7/4 > 1 and the exponent is positive,
   the function does not decay.
2.Because 4/5 < 1 and the exponent is negative,
    the function does not decay.
3. Because 8/7 > 1 and the exponent is negative,
    the function decays.
4. Because 9/2 > 1 and the exponent is positive,
    the function does not decay.

A composite plot the functions verifies the answer.

Answer: f(x)= \frac{3}{2} ( \frac{8}{7} )^{-x}

7 0
3 years ago
Read 2 more answers
I need help please fast
seraphim [82]

Answer:    6.46 is your answer

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Martin earned the following scores on his last five tests. 98, 78, 84, 75, 91 What is the mean absolute deviation of his scores?
rewona [7]
84 would be the answer.
6 0
3 years ago
Read 2 more answers
Prove: Every point of S=(0, 1) is an interior point of S.
myrzilka [38]

Answer:

Proved

Step-by-step explanation:

To prove that every point in the open interval (0,1) is an interior point of S

This we can prove by contradiction method.

Let, if possible c be a point in the interval which is not an interior point.

Then c has a neighbourhood which contains atleast one point not in (0,1)

Let d be the point which is in neighbourhood of c but not in S(0,1)

Then the points between c and d would be either in (0,1) or not in (0,1)

If out of all points say d1,d2..... we find that dn is a point which is in (0,1) and dn+1 is not in (0,1) however large n is.

Then we find that dn is a boundary point of S

But since S is an open interval there is no boundary point hence we get a contradiction. Our assumption was wrong.

Every point of S=(0, 1) is an interior point of S.

3 0
2 years ago
Please help need answers ASAP
Trava [24]

Answer:

0.527

Step-by-step explanation:

From the question,

The sequence is Geometry Progression (G.P)

Tₙ = arⁿ⁻¹.......................... Equation 1

Where n = number of term, a = first term, r = common ratio

Given: n = 7, a = 6, r = 2/3

Substitute these values into equation 1

T₇ = 6(2/3)⁷⁻¹

T₇ = 6(2/3)⁶

T₇ = 6(64/729)

T₇ = 384/729

T₇ = 0.527

Hence, the 7th term of the sequence is 0.527

4 0
2 years ago
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