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3 years ago

Please help me solve this?

Mathematics

1 answer:

jenyasd209 [6]3 years ago

6 0

Recipe is for 24 pieces, and required quantity is 16.

So the ratio is 16/24 = 2/3 means we need to reduce all ingredients to 2/3 of the given quantities. Not all recipes can be reduced prorata, but we will assume it can be done. A word of caution: the measurement of quantities will become awkward (like 3/8 of a teaspoon of salt, etc). The practical way is to make the full recipe, and invite friends to take home the remainder.

Here's the scaled recipe

Butter 1cup*3/4 = 3/4 cup

unsweetened chocolate = 4*3/4 = 3 squares

white sugar = 2 * 3/4 = 1 1/2 cups

eggs = 4 * 3/4 = 3 eggs

A.P. flour = 1 cup * 3/4 = 3/4 cups

vanilla extract = 1 * 3/4 = 3/4 teaspoons

salt = 1/2 * 3/4 = 3/8 teaspoons

chopped walnuts = 2 * 3/4 = 1 1/2 cups

For this reason, there is a trend to use weights (in grams) than volumes to make scaling recipes easier.

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Please help I dont get it

SpyIntel [72]

Answer:

B. 12cm

Step-by-step explanation:

We can double check by plugging in the number. However, the question trys to trick you, by giving you the diametre, but in order for the formula to work (V=4 /3πr^3), we will divide the diametre by 2 (12 ÷ 2= 6),

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Once we do that, we can double check, when you plug in your numbers your answer will be around 900 (904.78, to be exact, but the question asks for approximate diamtetre).

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Have a good day :)

3 0

3 years ago

Can someone please help me ?

Aliun [14]

Answer:

5x+3y=6

Step-by-step explanation:

Use the slope intercept form then rearrange

3 0

2 years ago

Find the greatest solution for x+y when x^2+y^2 = 7, x^3+y^3=10

damaskus [11]

Answer:

Step-by-step explanation:

set

$f(x,y)=x+y\\$

constrain:

$g(x,y)=x^2+y^2 = 7\\h(x,y)=x^3+y^3=10$

Partial derivatives:

$f_{x}=1\\f_{y} =1 \\g_{x}=2x \\g_{y}=2y\\h_{x}=3x^2 \\h_{y}=3y^2$

Lagrange multiplier:

$grad(f)=a*grad(g)+b*grad(h)\\$

$\left[\begin{array}{ccc}1\\1\end{array}\right]=a\left[\begin{array}{ccc}2x\\2y\end{array}\right]+b\left[\begin{array}{ccc}3x^2\\3y^2\end{array}\right]$

4 equations:

$1=2ax+3bx^2\\1=2ay+3by^2\\x^2+y^2=7\\x^3+y^3=10$

By solving:

$a=4/9\\b=-2/27\\x+y=4$

Second mathod:

Solve for x^2+y^2 = 7, x^3+y^3=10 first:

$x=\frac{1}{2} -\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} +\frac{\sqrt{13}}{2} \\x=\frac{1}{2} +\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} -\frac{\sqrt{13}}{2} \\x+y=-5\ or\ 1 \or\ 4$