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CaHeK987 [17]
3 years ago
9

How do you solve these types of questions?

Mathematics
1 answer:
Ahat [919]3 years ago
6 0
What questions? you didn't put any image up no one can answer your question because it's not a proper question.
You might be interested in
Why is the point ( 3, 4 ) the solution to this system of equations?<br> y= 1/3x + 3<br> y= 2x - 2
evablogger [386]

Answer:

The answer to your question is because it is a solution of each equation.

Step-by-step explanation:

Data

Point (3,4)

Equation l     y = 1/3x + 3

Equation ll     y = 2x - 2

Process

1.- To test is a point is solution to a system of equations,

-Substitute the point in both equations

Equation l            4 = 1/3(3) + 3

                            4 = 1 + 3

                            4 = 4               The point is a solution because

                                                     4 = 4

Equation ll            4 = 2(3) - 2

                             4 = 6 - 2

                             4 = 4               This point is a solution for the second

                                                     equation but not for the first one.

2.- Conclusion

As the point is a solution of each equation, it is a solution of the system.

6 0
3 years ago
Please answer answer now fast
iren [92.7K]

Answer:

fge or fgc

I think that because its kinda like the same

5 0
3 years ago
Read 2 more answers
How many sixteenths are there in 5/8​
nadya68 [22]

Answer:

10!

Step-by-step explanation:

Exuse me if im wrong :')

6 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

33.174666862=n

Hence answer is approx 33 years.

4 0
3 years ago
Read 2 more answers
Which describes the calculations that could be used to solve this problem? Lauren has 42 muffins to pack into boxes and bags. Sh
Marat540 [252]
The correct answer is D, 'Subtract 42 - 12. Then divide the difference by 3'. Lauren has 42 muffins, and she can put 12 of them into a box, leaving her with 30 muffins to fit into bags. If she can fit 3 muffins in each bag, and she wants to know how many bags she needs, then she needs to divide 30 by 3 to get 10. This means that she will need 10 bags of 3 muffins to completely pack 30 muffins. She started with 42, but 12 of those fit into the original box, so we don't have to worry about it.

Hope this helps!
6 0
3 years ago
Read 2 more answers
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