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AVprozaik [17]
3 years ago
7

If Oliver withdraws $285.10 from his savings account to make the purchase, how much would he have earned in interest on that amo

unt over five months? Round your answer to the nearest cent.
Mathematics
1 answer:
never [62]3 years ago
3 0
Oliver withdraws an amount of $285.10 from his saving account

His saving account earns 1.8% annually

The interest Oliver could have earned in five months:

Monthly interest = Annual interest ÷ 12
Monthly interest = 1.8% ÷ 12
Monthly interest = 0.018 ÷ 12 = 3/2000

After five months = Principle × (1 + interest)ⁿ
After five months = 285.10 × (1 + 0.018)⁵
After five months = 311.70

Interest earned = 311.70 - 285.10 = $26.60
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You have 12 coins, one of which is fake. The fake coin is indistinguishable from the rest except that it is heavier. Can you det
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Five friends Al, Bet, Cat, Don, and Ella live on the island where everyone is either a knight or a knave
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8 0
3 years ago
The coordinates of rhombus ABCD are A(–4, –2), B(–2, 6), C(6, 8), and D(4, 0). What is the area of the rhombus? Round to the nea
a_sh-v [17]
Check the picture below.

so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad 
C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2}
\\\\\\
AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\
-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad 
D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2}
\\\\\\
BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2}
\\\\\\
BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}

that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.

namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,

\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right]
\\\\\\
4[15]\implies 60

7 0
3 years ago
Read 2 more answers
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