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dalvyx [7]
3 years ago
13

If one pound of turkey costs 1.89 how many will x pounds cost

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
7 0
Hello There!

It would be 1.89 times how ever many pounds there are:
But in this case we don't know the number of pounds so it would be:
1.89(x)

Hope This Helps You!
Good Luck :) 

- Hannah ❤
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3 years ago
Please help me this is urgent.!
nata0808 [166]
V= 3.14• 5^2•4
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Sorry I couldn’t give a more detailed explanation but here is the original equation for volume of a cylinder:

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Hope this helps comment below for more questions :)
3 0
3 years ago
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Can i have help please?
alexandr1967 [171]

The means and medians are <em><u>not</u></em><em><u> </u></em> the same.

3 0
4 years ago
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if there are 2040 seats in a section of a stadium, 20 seats in the first row, and 4 more seats in each row, how many rows are th
Lorico [155]


2040 = n/2(20+(n-1)4)

4080 = n(20+4n-4)

4080= 20n +4n^2 -4n

1020 = 4n + n^2

n^2 +4n -1020 =0

use common formula (can't write out so just look at answers. sorry)

which gives answers of n=-34 and n=30. since n can only be positive, n=30 so there are 30 rows. I liked that challenge

5 0
3 years ago
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Let’s pretend that we want to give our employees in total a28% raise in the next two years, but we want to spread out aconsisten
Roman55 [17]

Solution

- The solution steps are given below:

\begin{gathered} \text{ Let the original amount be X} \\  \\ \text{ If we want a 28\% raise over the next two years, then it means that the amount will be:} \\ X+\frac{28}{100}X=1.28X \\  \\ \text{ If we want to give them a consistent raise each year, we can compute the scenario as follows:} \\ \text{ Let the percentage be }y \\  \\ X+y\%\text{ of }X=\text{ Salary after first year} \\  \\ (X+y\%\text{ of }X)+y\%\text{ of }(X+y\%\text{ of }X)\text{ = Salary after second year.} \\  \\ \text{ But we already know that the salary after second year is }1.28X \\ \text{ Thus, we can say:} \\ (X+y\%\text{ of }X)+y\%\text{ of }(X+y\%\text{ of }X)=1.28X \\  \\ \text{ Simplifying, we have:} \\ X+\frac{yX}{100}+\frac{y}{100}(X+\frac{yX}{100})=1.28X \\  \\ \text{ Divide through by }X \\ 1+\frac{y}{100}+\frac{y}{100}(1+\frac{y}{100})=1.28 \\  \\ \text{ Subtract 1 from both sides and expand the brackets} \\ \frac{y}{100}+\frac{y}{100}+(\frac{y}{100})^2=1.28-1=0.28 \\  \\ \frac{2y}{100}+(\frac{y}{100})^2=0.28 \\  \\ \text{ Multiply both sides by }100^2 \\ 200y+y^2=2800 \\  \\ \text{ Rewrite, we have:} \\ y^2+200y-2800=0 \\  \\ Solving\text{ the equation using the Quadratic formula, we have that:} \\ y=\frac{-200\pm\sqrt{200^2-4(-2800)}(1)}{2(1)} \\  \\ y=-213.137\text{   or   }13.137 \\  \\ \text{ Since the change in salary is an increase, thus, the rate }y\text{ has to be positive.} \\ \text{ Thus, } \\ y=13.137\approx13.14\% \\  \\  \end{gathered}

Final Answer

y = 13.14%

The screenshots of the solution are:

5 0
1 year ago
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