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Lelechka [254]
3 years ago
15

Larissa plans to bake at most 10 loaves of bread. She makes x loaves of banana bread that sell for $1.25 each and y loaves of nu

t bread that sell for $1.50 each. She hopes to make at least $24 in sales. Write and graph a system of inequalities for this situation. What does the graph show?
Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
3 0

Answer:

1) The system of inequalities are;

x × 1.25 + y × 1.5 ≥ 24

x + y ≤ 10

Please find attached the required graph of the two inequalities ,created with Microsoft Excel

2) The graph of inequality shows that Larissa cannot to make less than 10 loafs of bread and sell them for over $24 at the given prices of each loaf of bread

Step-by-step explanation:

1) The maximum number of loaves Larissa plans to make = 10

The price at which she sells each of x loaves of banana bread = $1.25

The price at which she sells each of y loaves of nut bread = $1.50

The total amount she hopes to make = $24

The system of inequalities are;

x × 1.25 + y × 1.5 ≥ 24

x + y ≤ 10

Making y the subject of both inequalities gives;

For the first inequality, we have,

y ≥ 24/1.5 - 1.25/1.5·x which gives;

y ≥ 16 - 5/6·x

For the second inequality, we have,

y ≤ 10 - x

Please find attached the required graph of the two inequalities

2) From the attached graph of inequality, created with Microsoft Excel, it shows that it is not possible for Larissa to make less than 10 loafs of bread and at the same time sell them for over $24 if the price of each banana bread is $1.25 and the price of each nut bread is $1.50.

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Help calculus module 8 DBQ<br><br> please show work
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1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of R(t) at the endpoints of each subinterval. The sum is then

\dfrac{R(0)+R(2)}2(2-0)+\dfrac{R(2)+R(3)}2(3-2)+\dfrac{R(3)+R(7)}2(7-3)+\dfrac{R(7)+R(8)}2(7-8)=\boxed{24.83}

which is measured in units of gallons, hence representing the amount of water that flows into the tank.

2. Since R is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval [a,b], it guarantees the existence of some c\in(a,b) such that

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R_{\rm avg}=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt

If doing this by hand, you can integrate by parts, setting

u=\ln(t^2+7)\implies\mathrm du=\dfrac{2t}{t^2+7}\,\mathrm dt

\mathrm dv=\mathrm dt\implies v=t

R_{\rm avg}=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_{t=0}^{t=8}-\int_0^8\frac{2t^2}{t^2+7}\,\mathrm dt\right)

For the remaining integral, consider the trigonometric substitution t=\sqrt 7\tan s, so that \mathrm dt=\sqrt 7\sec^2s\,\mathrm ds. Then

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\frac{7\tan^2s}{7\tan^2s+7}\sec^2s\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\tan^2s\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_{s=0}^{s=\tan^{-1}(8/\sqrt7)}

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\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt

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