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ch4aika [34]
3 years ago
7

PLEASE SOLVE AND CHECK. SHOW COMPLETE SOLUTION

Mathematics
1 answer:
vivado [14]3 years ago
3 0

Answer:

t \leq 25

Step-by-step explanation:

(\sqrt{2t+1}) ^{2} + (\sqrt{2t-1}) ^{2} \leq  10^{2}

=> 2t +1 + 2t -1 \leq 100

=> 4t \leq 100

=> t \leq 100/4

=> t \leq 25

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bazaltina [42]

Answer:

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3 years ago
Write the equation of the line, in slope-intercept form, that passes through the point (-8,-1) and that is parallel to x-3=0
nataly862011 [7]
Y=-5X which is the equation you use to figure out the answer
5 0
3 years ago
A species of beetles grows 32% every year. Suppose 100 beetles are released into a field. How many beetles will there be in 10 y
siniylev [52]

Given that a species of beetles grows 32% every year.

So growth rate is given by

r=32%= 0.32


Given that 100 beetles are released into a field.

So that means initial number of beetles P=100


Now we have to find about how many beetles will there be in 10 years.

To find that we need to setup growth formula which is given by

A=P(1+r)^n where A is number of beetles at any year n.

Plug the given values into above formula we get:

A=100(1+0.32)^n

A=100(1.32)^n


now plug n=10 years

A=100(1.32)^{10}=100(16.0597696605)=1605.97696605

Hence answer is approx 1606 beetles will be there after 20 years.


Now we have to find about how many beetles will there be in 20 years.

To find that we plug n=20 years

A=100(1.32)^{20}=100(257.916201549)=25791.6201549

Hence answer is approx 25791 beetles will be there after 20 years.



Now we have to find time for 100000 beetles so plug A=100000

A=100(1.32)^n

100000=100(1.32)^n

1000=(1.32)^n

log(1000)=n*log(1.32)

\frac{\log\left(10000\right)}{\log\left(1.32\right)}=n

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4 0
3 years ago
Read 2 more answers
Solve the following equation for the variable indicated<br> 15 = Зn + 6р, for n
Taya2010 [7]

Answer:

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Step-by-step explanation:

We need to solve the equation 15 = 3n + 6p for variable n

Solving:

15 = 3n + 6p

Subtract both sides by 6p

15-6p = 3n +6p -6p\\15-6p=3n

Switch sides of equality

3n=15-6p

Divide both sides by 3

\frac{3n}{3} =\frac{15-6p}{3}\\n=\frac{3(5-2p)}{3}\\n=5-2p

So, solving the equation 15 = 3n + 6p for variable n we get \mathbf{n=5-2p}

8 0
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BartSMP [9]
The answer is 70 the third option.
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