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kotegsom [21]
3 years ago
14

Math problem a recipe for a dessert requires 3 4/5 cups of strawberries and 1.5 cuos of yogurt . How many cups of strawberries s

hould be used if the recipe was tripled
Mathematics
1 answer:
Rashid [163]3 years ago
7 0

Answer:

The number of cups of strawberries to be used is 11 2/5 cups

Step-by-step explanation:

Here, we want to know how many cups of strawberries to be used given that the recipe was tripled.

From the question, we are told that the recipe required 3 4/5 cups of strawberries.

Now, by tripling the recipe, it means we will be tripling the individual cups too.

So the number of cups of strawberries to be used will be ;

3 4/5 * 3

3 4/5 as improper fraction is 19/5

so we have 19/5 * 3 = 57/5 = 11 2/5

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. If her allowance had stayed the same, $9.00 a month, how many carnival tickets could she bu
jarptica [38.1K]

Answer:

a) number of tickets she could buy in a month with her allowance in 2008 = 4 tickets

b) number of tickets she could buy in a month with the same allowance in 2012 = 2 tickets

c) $5, an increase of 55.56%

d) $2, a 100% increase.

e) Hallie was able to buy more tickets in 2008 than in 2012.

f) $16

Step-by-step explanation:

The complete question is presented in the attached image to this answer.

a) Halie's allowance monthly = $9

Price of a ticket in 2008 = $2

number of tickets she could buy in a month with her allowance = (9/2) = 4.5 = 4 tickets (since there are no half tickets)

b) Halie's allowance monthly = $9

Price of a ticket in 2012 = $4

number of tickets she could buy in a month with her allowance = (9/4) = 2.25 = 2 tickets (since there are no quarter tickets)

c) In 2012, Hallie's allowance was $14.00 per month. How much did her monthly allowance increase between 2008 and 2012?

Her monthly allowance in 2008 = $9

Her monthly allowance in 2012 = $14

The increase = new allowance - old allowance = 14 - 9 = $5

In percentage terms,

% increase = 100% × (new - old)/(old)

% increase = 100% × (14 - 9)/9 = 55.56%

d) How much more did a carnival ticket cost in 2012 than it did in 2008?

Price of a ticket on 2008 = $2

Price of a ticket in 2012 = $4

Increase in price = 4 - 2 = $2

% increase = 100% × (4 - 2)/2 = 100%

e) Was Hallie able to buy more carnival tickets in 2008 or in 2012 with one month's allowance?

Hallie was able to buy more tickets in 2008 than in 2012.

f) What would Hallie's allowance need to be in 2012 in order for her to be able to buy as many carnival tickets as she could in 2008?

Hallie could buy 4 tickets in 2008.

Price of a ticket in 2012 = $4.

Price of 4 tickets in 2012 = 4 × $4 = $16

Hope this Helps!!!!

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Answer:

S=-2

Step-by-step explanation:

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3 years ago
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Check the picture below.

so the perimeter of the polygon is the sum of all its sides, namely, AB + BC + CD + DA.

now, let's check how long each side is,

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&A&(~{{ -6}} &,&{{ -4}}~) &#10;%  (c,d)&#10;&B&(~{{ -3}} &,&{{ 6}}~)&#10;\end{array}&#10;\\\\\\&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\&#10;-------------------------------\\\\&#10;AB=\sqrt{[-3-(-6)]^2+[6-(-4)]^2}&#10;\\\\\\&#10;AB=\sqrt{(-3+6)^2+(6+4)^2}&#10;\\\\\\&#10;AB=\sqrt{3^2+10^2}\implies \boxed{AB=\sqrt{109}}\\\\&#10;-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&B&(~{{ -3}} &,&{{6}}~) &#10;%  (c,d)&#10;&C&(~{{ 4}} &,&{{ 0}}~)&#10;\end{array}&#10;\\\\&#10;-------------------------------\\\\&#10;BC=\sqrt{[4-(-3)]^2+[0-6]^2}\implies BC=\sqrt{(4+3)^2+(0-6)^2}&#10;\\\\\\&#10;BC=\sqrt{7^2+(-6)^2}\implies \boxed{BC=\sqrt{85}}\\\\&#10;-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&C&(~{{ 4}} &,&{{0}}~) &#10;%  (c,d)&#10;&D&(~{{ 2}} &,&{{ -1}}~)&#10;\end{array}&#10;\\\\&#10;-------------------------------\\\\&#10;CD=\sqrt{(2-4)^2+(-1-0)^2}\implies CD=\sqrt{(-2)^2+(-1)^2}&#10;\\\\\\&#10;\boxed{CD=\sqrt{5}}\\\\&#10;-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&D(~{{ 2}} &,&{{-1}}~) &#10;%  (c,d)&#10;&A&(~{{ -6}} &,&{{ -4}}~)&#10;\end{array}\\\\&#10;-------------------------------\\\\&#10;DA=\sqrt{[-6-2]^2+[-4-(-1)]^2}\\\\\\ DA=\sqrt{(-6-2)^2+(-4+1)^2}&#10;\\\\\\&#10;DA=\sqrt{(-8)^2+(-3)^2}\implies \boxed{DA=\sqrt{73}}

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