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3 years ago

12

Lily uses 1/3 pints of milk and 2 bananas to prepare a milkshake. If she uses 2/3 pints of milk, how many bananas would she need

to make the milkshake?

1 answer:

telo118 [61]3 years ago

8 0

Answer:

4 bananas

Step-by-step explanation:

She uses 4 bananas because if 2 are needed to prepare 1/3 pints for 2 then 2/3 u need 4

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What is the answer of 2 divide by 3/4?

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Answer: 2.66666666667

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There are two numbers. The first number minus the second number is 15. One-third of the sum of the numbers is one quarter of the

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Answer:

So the numbers are 12 and -3.

Step-by-step explanation:

In order to solve this problem we will attribute variables to the numbers, the first one will be "x" and the second one will be "y". From the first sentence we know that the subtraction of the two numbers is equal to 15, so we have:

x - y = 15

Then the problem states that one-third of the sum of the number is equal to one quarter of the first number, so we have:

(1/3)*(x+y) = x/4

Since we now have two equations and two variables we can solve for x and y. From the first equation we have:

y = x - 15

Using this expression for the value of y in the second equation:

(1/3)*(x + x - 15) = x/4

(1/3)*(2*x - 15) = x/4

2*x - 15 = 3*x/4

2*x - 3*x/4 = 15

(8*x - 3*x)/4 = 15

5*x/4 = 15

5*x = 60

x = 60/5 = 12

y = x - 15 = 12 - 15 = -3

So the numbers are 12 and -3.

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If sin A is negative and cos A is positive, what

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Answer:

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Step-by-step explanation:

sinA < 0 in quadrants 3 and 4

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A rectangular swimming pool has a length twice as long as its width. The pool has a sidewalk around it that is 2 feet wide. Writ

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Answer:

$A_p=2w^2$

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Step-by-step explanation:

Functions and Geometry

Sometimes we need to express areas, volumes, and distances in terms of unknown quantities (or variables). When we can find such expressions, they can be referred to as functions. Let's recall the area of a rectangle is A=wl, being w its wide and l its length

Please refer to the image below. The length of the swimming pool is l and its width is w. Its area is

$A_p=lw$

We know the length is twice as long as its width, l=2w, so

$A_p=(2w)w=2w^2$

$A_p=2w^2$

The sidewalk is 2 feet wide. The length of the pool and its sidewalk is

$l'=l+4$

Similarly, the width of the pool+sidewalk is

$w'=w+4$

The area of the pool+sidewalk is

$A'=l'w'=(l+4)(w+4)=lw+4l+4w+16$

We know that l=2w, so

$A'=(2w)w+4(2w)+4w+16$

$A'=2w^2+12w+16$

The area of the sidewalk alone is the subtraction of both areas

$A_s=A'-A_p=2w^2+12w+16-2w^2$

Simplifying

$A_s=12w+16$

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