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fiasKO
3 days ago
10

6. Charlie can spend up to $8 on lunch. He wants to buy a tuna sandwich, a bottle of

Mathematics
1 answer:
PIT_PIT [9.1K]3 days ago
3 0

Response:

Charlie can purchase a maximum of 0.375 pounds of potato salad

Detailed explanation:

Refer to the figure provided for improved clarity on the issue

Let

a ----> the price of one tuna sandwich

b ----> the price of a bottle of apple juice

c ----> the cost per pound of potato salad

x ----> pounds of potato salad

we have

a=\$4.25

b=\$2.25

c=\$4.00/lb

he intends to buy a tuna sandwich, a bottle of apple juice, and x pounds of potato salad, and his spending limit is $8

The corresponding inequality for this scenario is

a+b+cx \leq 8

substituting the known values

4.25+2.25+4.00x \leq 8

Solving for x

Combine similar terms

6.50+4.00x \leq 8

Decrease 6.50 from both sides

4.00x \leq 8-6.50

4.00x \leq 1.50

Divide both sides by 4

x \leq 1.50/4.00

x \leq 0.375\ lbs

thus

Charlie can acquire a maximum of 0.375 pounds of potato salad

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Response:

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Detailed explanation:

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16 days ago
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At what point of the curve y = cosh(x does the tangent have slope 2?
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The hyperbolic cosine function (cosh) is defined as
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4 days ago
12x+7<−11 AND5x−8≥4012
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Answer:

No solution

Step-by-step explanation:

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Handle each inequality separately.

12x+7              Utilizing the subtraction property of inequalities

12x

12x

x

x

and

5x-8\geq 40               Utilizing the addition property of inequalities

5x\geq 40+8

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x\geq \dfrac{48}{5}

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Refer to the attached image for the number line representation.

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\bf \qquad \qquad \textit{Annual Yield Formula}
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~~~~~~~~~~~~\textit{4.0784\% compounded monthly}\\\\
~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
\\\\
\begin{cases}
r=rate\to 4.0784\%\to \frac{4.0784}{100}\to &0.040784\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{monthly, thus twelve}
\end{array}\to &12
\end{cases}
\\\\\\
\left(1+\frac{0.040784}{12}\right)^{12}-1\\\\
-------------------------------\\\\
~~~~~~~~~~~~\textit{4.0798\% compounded semiannually}\\\\


\bf ~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1
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\begin{cases}
r=rate\to 4.0798\%\to \frac{4.0798}{100}\to &0.040798\\
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\begin{array}{llll}
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\bf ~~~~~~~~~~~~\textit{4.0730\% compounded daily}\\\\
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Subsequently

The remaining depth needed for filling with water = (18-3) inches

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