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ira [324]
2 years ago
6

Given f(x)=3^x-8, what is the value of f(10)

Mathematics
1 answer:
Westkost [7]2 years ago
8 0

Answer:

f(x) =  {3}^{(x)}  - 8 \\ f(10) =  {3}^{(10)} - 8  \\  = 59041

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Calculate the perimeter of a square length of whose side is 40cm​
Verizon [17]

Answer:

160 cm

Step-by-step explanation:

a square has 4 equal sides. if one of the sides is 40 cm, then all of the other sides are 40 cm as well. to get a perimeter, you add up the lengths of all the sides. if there are 4 sides of 40 cm, you multiply 40 by 4 to get 160, finally add the unit to get 160 cm.

8 0
3 years ago
Please help me please please please
algol13

Answer:

20

Step-by-step explanation:

The lower quartile Q₁ is the value at the left side of the box, that is

Q₁ = 20

5 0
3 years ago
Read 2 more answers
The container that holds the water for the football team is
aliya0001 [1]
15 gallons
hope this helps. not really sure what you mean by that.
3 0
3 years ago
Read 2 more answers
What is 4 to the power of x minus 3 =18
mote1985 [20]
3 to the power of 4:

=34
x
n
=
3
4

=3⋅3⋅3⋅3
=
3
⋅
3
⋅
3
⋅
3

=81
=
81

For example, 3 to the power of -4:

=3−4
x
n
=
3
−
4

=134
=
1
3
4

=13⋅3⋅3⋅3
=
1
3
⋅
3
⋅
3
⋅
3

=181
=
1
81

=0.012346
7 0
2 years ago
Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x) = x/6x^2 +
timama [110]

Looks like your function is

f(x)=\dfrac x{6x^2+1}

Rewrite it as

f(x)=\dfrac x{1-(-6x^2)}

Recall that for |x|, we have

\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n

If we replace x with -6x^2, we get

f(x)=\displaystyle x\sum_{n=0}^\infty\frac(-6x^2)^n=\sum_{n=0}^\infty (-6)^n x^{2n+1}

By the ratio test, the series converges if

\displaystyle\lim_{n\to\infty}\left|\frac{(-6)^{n+1} x^{2(n+1)+1}}{(-6)^n x^{2n+1}}\right|=6|x^2|\lim_{n\to\infty}1=6|x|^2

Solving for x gives the interval of convergence,

|x|^2

We can confirm that the interval is open by checking for convergence at the endpoints; we'd find that the resulting series diverge.

3 0
2 years ago
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