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sdas [7]
2 years ago
7

A recipe for pie please help me please

Mathematics
2 answers:
Natalija [7]2 years ago
4 0
I highly suggest you buy the crust pre made because I made it once and it’s so hard to do and never comes out as good as you want it to
motikmotik2 years ago
3 0

Answer:

Apple Pie (This recipe is from TasteofHome I did not make it)

Step-by-step explanation:

Ingredients

1/2 cup sugar

1/2 cup packed brown sugar

3 tablespoons all-purpose flour

1 teaspoon ground cinnamon

1/4 teaspoon ground ginger

1/4 teaspoon ground nutmeg

6 to 7 cups thinly sliced peeled tart apples

1 tablespoon lemon juice

Dough for double-crust pie

1 tablespoon butter

1 large egg white

Additional sugar

Directions

Preheat oven to 375°. In a small bowl, combine sugars, flour and spices; set aside. In a large bowl, toss apples with lemon juice. Add sugar mixture; toss to coat.

On a lightly floured surface, roll one half of dough to a 1/8-in.-thick circle; transfer to a 9-in. pie plate. Trim even with rim. Add filling; dot with butter. Roll remaining dough to a 1/8-in.-thick circle. Place over filling. Trim, seal and flute edge. Cut slits in top. Beat egg white until foamy; brush over crust. Sprinkle with sugar. Cover edge loosely with foil.

Bake 25 minutes. Remove foil; bake until crust is golden brown and filling is bubbly, 20-25 minutes longer. Cool on a wire rack.

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Artist 52 [7]
Check the picture below.  Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm².  Also, recall, the base is a square, thus, length = width = x.

\bf \textit{volume of a rectangular prism}\\\\&#10;V=lwh\quad &#10;\begin{cases}&#10;l = length\\&#10;w=width\\&#10;h=height\\&#10;-----\\&#10;w=l=x&#10;\end{cases}\implies V=xxh\implies \boxed{V=x^2h}\\\\&#10;-------------------------------\\\\&#10;\textit{surface area}\\\\&#10;S=4xh+x^2\implies 300=4xh+x^2\implies \cfrac{300-x^2}{4x}=h&#10;\\\\\\&#10;\boxed{\cfrac{75}{x}-\cfrac{x}{4}=h}\\\\&#10;-------------------------------\\\\&#10;V=x^2\left( \cfrac{75}{x}-\cfrac{x}{4} \right)\implies V(x)=75x-\cfrac{1}{4}x^3

so.. that'd be the V(x) for such box, now, where is the maximum point at?

\bf V(x)=75x-\cfrac{1}{4}x^3\implies \cfrac{dV}{dx}=75-\cfrac{3}{4}x^2\implies 0=75-\cfrac{3}{4}x^2&#10;\\\\\\&#10;\cfrac{3}{4}x^2=75\implies 3x^2=300\implies x^2=\cfrac{300}{3}\implies x^2=100&#10;\\\\\\&#10;x=\pm10\impliedby \textit{is a length unit, so we can dismiss -10}\qquad \boxed{x=10}

now, let's check if it's a maximum point at 10, by doing a first-derivative test on it.  Check the second picture below.

so, the volume will then be at   \bf V(10)=75(10)-\cfrac{1}{4}(10)^3\implies V(10)=500 \ cm^3

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