Like all problems that involve images within the question, we should definitely try to draw this out. In the picture above, I have done this.
Now, we can see that this is just a simple proportion problem. For every 2.5 cm of height of the flower, we are 2 cm from the opening, or aperture. For every 20 cm of height, how far are we? We can set up the problem like this:
20 ............2.5
-------- = ---------
...x ............. 2
where x is the unknown distance to the aperture from the flower. Now, we just need to get x by itself. A typical way of solving something like this is by doing "butterfly multiplication" which is really just a shortcut haha. Anyway, I can rewrite that equation ^ as:
20×2 = 2.5 × x
Then, to solve for x, we would divide both sides by 2.5. (If you don't know why that is, please let me know and I'll elaborate).
We would then have:
20×2
------- = x
2.5
Which then simplifies to:
x = 16
Try using the same logic for your second question, and if you get stuck, I'd be happy to help! please let me know if any of this doesn't make sense. :)
Answer:
For a monthly cost of at least $7 and at most $8, you can have between 100 and 110 calling minutes.
Step-by-step explanation:
The problem states that the monthly cost of a celular plan is modeled by the following function:

In which C(x) is the monthly cost and x is the number of calling minutes.
How many calling minutes are needed for a monthly cost of at least $7?
This can be solved by the following inequality:






For a monthly cost of at least $7, you need to have at least 100 calling minutes.
How many calling minutes are needed for a monthly cost of at most 8:






For a monthly cost of at most $8, you need to have at most 110 calling minutes.
For a monthly cost of at least $7 and at most $8, you can have between 100 and 110 calling minutes.
e^x=170
because you subtract like terms (or add, it depends) then simplify (im bad at explaining)
0.270 repeated. I used long division and a calculator. hoped this helped.
Answer:
1) The solve by graphing will the preferred choice when the equation is complex to be easily solved by the other means
Example;
y = x⁵ + 4·x⁴ + 3·x³ + 2·x² + x + 3
2) Solving by substitution is suitable where we have two or more variables in two or more (equal number) of equations
2x + 6y = 16
x + y = 6
We can substitute the value of x = 6 - y, into the first equation and solve from there
3) Solving an equation be Elimination, is suitable when there are two or more equations with coefficients of the form, 2·x + 6·y = 23 and x + y = 16
Multiplying the second equation by 2 and subtracting it from the first equation as follows
2·x + 6·y - 2×(x + y) = 23 - 2 × 16
2·x - 2·x + 6·y - 2·y = 23 - 32
0 + 4·y = -9
4) An example of a linear system that can be solved by all three methods is given as follows;
2·x + 6·y = 23
x + y = 16
Step-by-step explanation: