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Afina-wow [57]
3 years ago
7

Raquel went to Wellers department stores one day sale she bought two blouses for $38 each and a pair of shoes for $49 she also w

anted to buy some jewelry each item of Jewelry was bargain price at $11.50 each if she bought $171 with her how many pieces of jewelry could she buy
Mathematics
2 answers:
Svetlanka [38]3 years ago
4 0
She can buy 4 pieces of jewelry she boght 2 blouses for $38 38+38=76 and buys a pair of shoes for $49 76+49=125 so 171-125=46 so then just divide 46 and 11.5 which equals 4.
Oxana [17]3 years ago
3 0
38+38+49+11.5x≤171
125+11.5x≤171
11.5x≤46
x≤4

Answer:She could buy 4 pieces of jewlery

Hope this helps:)
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What is the remainder when 864 is devided by 31? 6 27 4 ther is no remainder
avanturin [10]
Okay. So when you solve 864/31 on a sheet of paper and you divide it properly, the remainder you get should be 27. The answer is B: 27.

6 0
3 years ago
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. The time of a clock as seen in a mirror is 3:15. The correct time is​
Oksana_A [137]

Answer:

3:15 cause a mirror cant change time

Step-by-step explanation:

go in ur mirror and look at the time did it somehow get darker or lighter???

8 0
3 years ago
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A person invests $4000 at 2% interest compounded annually for 4 years and then invests the balance (the $4000 plus the interest
faltersainse [42]
\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4000\\
r=rate\to 2\%\to \frac{2}{100}\to &0.02\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &4
\end{cases}
\\\\\\
A=4000\left(1+\frac{0.02}{1}\right)^{1\cdot 4}\implies A=4000(1.02)^4\implies A\approx 4329.73

then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4329.73\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &7
\end{cases}
\\\\\\
A=4329.73\left(1+\frac{0.08}{1}\right)^{1\cdot 7}\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396

add both amounts, and that's her investment for the 11 years.
7 0
3 years ago
Please Show the work
Svetradugi [14.3K]

the answer is a) because its the numerator is bigger than the denominator

8 0
3 years ago
Consider the line y=-3x+3.
emmainna [20.7K]

Answer:

y= 3x+17; y= -\frac13x +\frac{11}3

Step-by-step explanation:

First problem. If you want a parallel to a given line, you keep the slope.

Then we use the point-slope form of a line

y-y_0=m(x-x_0) and we plug in there everything we need.

y-5=3(x+4) \rightarrow y=3x+12+5\\y=3x+17

The second is quite similar. This time we want the perpendicular. It means that the product of the slopes has to be -1.

3\cdot m = -1 \rightarrow m=-\frac13

At this point we have everything, let's replace and write down the line in a better looking form

y-5=-\frac13(x+4) \rightarrow y= -\frac13 x -\frac43 +5\\y= -\frac13x -\frac43 +\frac{15}3 \rightarrow y= -\frac13x +\frac{11}3

7 0
2 years ago
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