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vladimir2022 [97]

3 years ago

8

A recipe calls for chocolate chip in a ratio of 1 cup of white chocolate chips to 2 cups of white chocolate chips how many milk

chocolate chips were added if there where 6 cups of chocolate chips total?

1 answer:

HACTEHA [7]3 years ago

4 0

You have 1:2 and then you have _:6 so you would divide 6by 2 and get 3. 3:6

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HELP PLEASEEEE!!

emmasim [6.3K]

31.0 degrees

TanX wich is 30/50

same thing as 30 divided by 50

7 0

2 years ago

Two teams of interns are wrapping donated gifts at a hospital. if the teams work alone, team a can wrap all of the gifts in 7 ho

romanna [79]

Together it would take 3 hours.
Team A can wrap all gifts in 7 hours; thus they can wrap 1/7 of the gifts in 1 hour.

Team B can wrap all gifts in 5 hours; thus they can wrap 1/5 of the gifts in 1 hour.

1/7h + 1/5h = 1, where h is the number of hours; it equals 1 because it is 100% of the gifts;

Find a common denominator. 35 is the first thing 7 and 5 will both divide into. Convert the fractions:
5/35h + 7/35h = 1
12/35h = 1

Divide both sides by 12/35:
12/35h ÷ 12/35 = 1 ÷ 12/35
h = 1/1 ÷ 12/35
h = 1/1 × 35/12
h = 35/12
h = 2.9
h ≈ 3

4 0

3 years ago

Suppose a rectangular pasture is to be constructed using 1 2 linear mile of fencing. The pasture will have one divider parallel

timama [110]

Answer:

$\displaystyle A=\frac{1}{192}$

Step-by-step explanation:

<u>Maximization With Derivatives</u>

Given a function of one variable A(x), we can find the maximum or minimum value of A by using the derivatives criterion. If A'(x)=0, then A has a probable maximum or minimum value.

We need to find a function for the area of the pasture. Let's assume the dimensions of the pasture are x and y, and one divider goes parallel to the sides named y, and two dividers go parallel to x.

The two divisions parallel to x have lengths y, thus the fencing will take 4x. The three dividers parallel to y have lengths x, thus the fencing will take 3y.

The amount of fence needed to enclose the external and the internal divisions is

$P=4x+3y$

We know the total fencing is 1/2 miles long, thus

$\displaystyle 4x+3y=\frac{1}{2}$

Solving for x

$\displaystyle x=\frac{\frac{1}{2}-3y}{4}$

The total area of the pasture is

$A=x.y$

Substituting x

$\displaystyle A=\frac{\frac{1}{2}-3y}{4}.y$

$\displaystyle A=\frac{\frac{1}{2}y-3y^2}{4}$

Differentiating with respect to y

$\displaystyle A'=\frac{\frac{1}{2}-6y}{4}$

Equate to 0

$\displaystyle \frac{\frac{1}{2}-6y}{4}=0$

Solving for y

$\displaystyle y=\frac{1}{12}$

And also

$\displaystyle x=\frac{\frac{1}{2}-3\cdot \frac{1}{12}}{4}=\frac{1}{16}$

Compute the second derivative

$\displaystyle A''=-\frac{3}{2}.$

Since it's always negative, the point is a maximum

Thus, the maximum area is

$\displaystyle A=\frac{1}{12}\cdot \frac{1}{16}=\frac{1}{192}$

6 0

3 years ago

ipn [44]

Answer:

a=9

b=9

c=18

Step-by-step explanation:

c(a+b)+a+b=36

2a+2b=36

a+b=18

c=18

4 0

2 years ago

Graph the numbers that are solutions to both inequalities below.

aleksklad [387]

Answer:

3x>12 2x/3 ≤ 6

3x>12, x = 4 2x/3, x = 2 ≤ 6

3 0

3 years ago