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ahrayia [7]
3 years ago
14

Sheila eats 3/4 of a bag of baby carrots each week. How many bags of baby carrots does she eat in 6 weeks?

Mathematics
2 answers:
DedPeter [7]3 years ago
8 0
She would have eaten 4.5 bags
Katyanochek1 [597]3 years ago
6 0
Multiply 6 by 3/4:
\frac{3}{4} \times6= \frac{3}{2}\times3=\frac{9}{2}
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I've been having problems with this for a while...
Rina8888 [55]
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. :)

3 0
4 years ago
A basic cellular phone plan costs $4 per month for 70 calling minutes. Additional time costs $0.10 per minute. The formula C= 4+
Harrizon [31]

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:

C(x) = 4 + 0.10(x-70)

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:

C(x) \geq 7

4 + 0.10(x - 70) \geq 7

4 + 0.10x - 7 \geq 7

0.10x \geq 10

x \geq \frac{10}{0.1}

x \geq 100

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:

C(x) \leq 8

4 + 0.10(x - 70) \leq 8

4 + 0.10x - 7 \leq 8

0.10x \leq 11

x \leq \frac{11}{0.1}

x \leq 110

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.

8 0
3 years ago
9e^x-240=6e^x+270 what is the solution
zmey [24]

e^x=170

because you subtract like terms (or add, it depends) then simplify (im bad at explaining)

4 0
3 years ago
What the answer to lesson 5.4
Anastaziya [24]
0.270 repeated. I used long division and a calculator. hoped this helped.


6 0
3 years ago
What situations would solve by graphing be your preferred choice? Give an example.
Damm [24]

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:

4 0
3 years ago
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