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Norma-Jean [14]
3 years ago
12

Please help me with 9&10

Mathematics
1 answer:
Andrew [12]3 years ago
3 0
This is problem #9. I'm not sure how to solve #10, although the solution may be found in a similar way.

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devin began running a month ago to get back in shape.the first day he ran .5 miles. each day after that he ran 10% more than the
SpyIntel [72]
The formula for the sum of the a finite geometric series is this:
Sn = a((1-r^n) / (1-r))

The given values are:
n = 30
a = 0.5 mi
r = 1.10 (10% more each day)

S30 = 0.5 ((1-1.10^30) / (1-1.10))
S30 = 82.25 mi

After 30 days, he will have traveled 82.25 miles.
3 0
4 years ago
Can you write to me explaining this please
Mnenie [13.5K]

Answer:

In addition, from the response shown, using a graphical calculator brings the following benefits:

1) You can write the system of linear equations as big as you want. This is: systems 3 * 3, 4 * 4, 5 * 5.

2) The response to systems of equations greater than 2 * 2 can be complicated when you graph the solution, therefore, the graphing calculator can be much more efficient in these cases.

3) You can write the linear equations in any way. Resolving by hand you should probably rewrite the system of equations to find the solution.

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Define the word evaluate. in complete sentences
il63 [147K]

Answer:

A woman evaluates a gemstone to determine it's value. Licensed from iStockPhoto. verb. To evaluate is defined as to judge the value or worth of someone or something.

I hope It's right. Sorry if it isn't.

Have a wonderful day/night :)

8 0
3 years ago
A particle moves along the x-axis so that at time t its position is given by s(t) = (t + 1)(t - 3)3, t > 0. /p>For what va
Zepler [3.9K]

Given the position function, the velocity function is obtained by taking the derivative:

s(t)=(t+1)(t-3)^3\implies v(t)=(t-3)^3+3(t+1)(t-3)^2

The velocity is increasing its own derivative is positive, so we also have to find the acceleration by taking another derivative:

a(t)=4(t-3)^3+3(t-3)^2+6(t+1)(t-3)

To find when a(t)>0, we first need to know where a(t)=0:

4(t-3)^3+3(t-3)^2+6(t+1)(t-3)=(t-3)\bigg(4(t-3)^2+3(t-3)+6(t+1)\bigg)=(t-3)(4t^2-15t+33)=0

The quadratic factor is always positive (its discriminant is negative), which leaves one solution at t=3. To either side of t=3, we have, for instance,

a(2)=-12

a(4)=36>0

which indicates that v(t) is increasing for t>3, making the answer A.

5 0
3 years ago
If h(x) = 2x² - 5 find h(4x²)​
nydimaria [60]

Answer:

8x^4-5

Step-by-step explanation:

8x^4-5

Just substitute 4x^2 into x

4 0
3 years ago
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