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1 year ago

7

Priscilla’s grandmother’s fruit salad recipe calls for one part apple, one part orange, four parts strawberry, two parts cherry,

and three parts grape. Priscilla uses the same measuring cup to measure all of the fruit, so one part is equal to one cup of diced fruit. In this exercise, you will compare the quantities in the recipe to understand ratios better.

1 answer:

Kryger [21]1 year ago

8 0

There are 11 parts of fruit: 4-strawberry, 3- grape, 2, cherry, 1- apple, 1-orange

4/11 from the entine quantity of fruits- strawberry
3/11-grapes
2/11- cherries
1/11- apples
1/11- oranges } ⇒ the quantity of oranges and apples are equal

The biggest part of the fruit- 4/11 - are strawberries
4/11>3/11>2/11>1/11
the smallest part of the fruit- apples and oranges

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Step-by-step explanation:

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If a linear system has four equations and seven variables, then it must have infinitely many solutions.

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Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.Tiles

Ugo [173]

By definition, the average rate of change is given by:

$AVR = \frac{f(x2)-f(x1)}{x2-x1}$

We evaluate each of the functions in the given interval.

We have then:

For f (x) = x ^ 2 + 3x:

Evaluating for x = -2:

$f (-2) = (-2) ^ 2 + 3 (-2)\\f (-2) = 4 - 6\\f (-2) = - 2$

Evaluating for x = 3:

$f (3) = (3) ^ 2 + 3 (3)\\f (3) = 9 + 9\\f (3) = 18$

Then, the AVR is:

$AVR = \frac{18-(-2)}{3-(-2)}$

$AVR = \frac{18+2}{3+2}$

$AVR = \frac{20}{5}$

$AVR = 4$

For f (x) = 3x - 8:

Evaluating for x =4:

$f (4) = 3 (4) - 8\\f (4) = 12 - 8\\f (4) = 4$

Evaluating for x = 5:

$f (5) = 3 (5) - 8\\f (5) = 15 - 8\\f (5) = 7$

Then, the AVR is:

$AVR = \frac{7-4}{5-4}$

$AVR = \frac{3}{1}$

$AVR = 3$

For f (x) = x ^ 2 - 2x:

Evaluating for x = -3:

$f (-3) = (-3) ^ 2 - 2 (-3)\\f (-3) = 9 + 6\\f (-3) = 15$

Evaluating for x = 4:

$f (4) = (4) ^ 2 - 2 (4)\\f (4) = 16 - 8\\f (4) = 8$

Then, the AVR is:

$AVR = \frac{8-15}{4-(-3)}$

$AVR = \frac{-7}{4+3}$

$AVR = \frac{-7}{7}$

$AVR = -1$

For f (x) = x ^ 2 - 5:

Evaluating for x = -1:

$f (-1) = (-1) ^ 2 - 5\\f (-1) = 1 - 5\\f (-1) = - 4$

Evaluating for x = 1:

$f (1) = (1) ^ 2 - 5\\f (1) = 1 - 5\\f (1) = - 4$

Then, the AVR is:

$AVR = \frac{-4-(-4)}{1-(-1)}$

$AVR = \frac{-4+4}{1+1}$

$AVR = \frac{0}{2}$

$AVR = 0$

Answer:

from the greatest to the least value based on the average rate of change in the specified interval:

f(x) = x^2 + 3x interval: [-2, 3]

f(x) = 3x - 8 interval: [4, 5]

f(x) = x^2 - 5 interval: [-1, 1]

f(x) = x^2 - 2x interval: [-3, 4]

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