First, convert the 25% to a real mathematical number. For percents, this is always done by dividing the 25% by 100%, or 25% / 100% = 0.250.
Second, find out what 25% of $100 is. This is the amount of the sale discount. This is always found by mulitplying 0.250 by the item's cost $100, like this:
0.250 x $100 = $25.00.
So for this sale, you'll save $25.00 on this item.
This means, the cost of the item to you is
$100 - $25.00 = $75.00.
Alternatively, you can think about it this way. The item is 25% off. This means you'll pay 75.000% of the total cost (100% - 25% = 75.000%).
Now what's 75.000% of the total cost?
0.750 x $100 = $75.00.
Just like the result above, the sale price on the item is $75.00.
Hope this helps you on your assignment! :)
Answer:
<h3>a)a+c=150,10.25a+7.75c=1470</h3>
Step-by-step explanation:
<h3>to understand this</h3><h3>you need to know about:</h3>
- system of linear equation
<h3>let's solve:</h3>
according to the first condition
according to the second condition
Answer:
the answer is D
Step-by-step explanation:
I don't know how to explain it very well but the line starts at 2 then ends at 8
<span>2(4/2)^2 - 15 + 6
= </span><span>2(2)^2 - 15 + 6
= </span><span>2(4) - 15 + 6
= 8 - 15 + 6
= -1</span>
let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B6%7D%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B6%5E2-5%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B36-25%7D%5Cimplies%20%5Cpm%20%5Csqrt%7B11%7D%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
