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LiRa [457]
2 years ago
5

A recipe calls for 14cup of brown sugar for every 23cup of white sugar. How many cups of brown sugar are required for every cup

of white sugar? 1Points A 18cup of brown sugar B 38cup of brown sugar C 14cup of brown sugar D 34cup of brown sugar
Mathematics
1 answer:
photoshop1234 [79]2 years ago
6 0

Note: Consider all the numeric values are in the fraction form. Like 14\to \dfrac{1}{4}, 23\to \dfrac{2}{3} etc.

Given:

A recipe calls for \dfrac{1}{4} cup of brown sugar for every \dfrac{2}{3} cup of white sugar.

To find:

The number of cups of brown sugar that are required for every cup of white sugar.

Solution:

We know that,

Required brown sugar for \dfrac{2}{3} cup of white sugar =  \dfrac{1}{4} cup

Using this, we get

Required brown sugar for 1 cup of white sugar =  \dfrac{\dfrac{1}{4}}{\dfrac{2}{3}} cup

                                                                              =\dfrac{1}{4}\times \dfrac{3}{2} cup

                                                                              =\dfrac{3}{8} cup

So, \dfrac{3}{8} cup of brown sugar is required for every cup of white sugar.

Therefore, the correct option is B.

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I need help pleases I wil give brainyless
erica [24]

Answer: Let's start by writing down coordinates of all points:

A(0,0,0)

B(0,5,0)

C(3,5,0)

D(3,0,0)

E(3,0,4)

F(0,0,4)

G(0,5,4)

H(3,5,4)

a.) When we reflect over xz plane x and z coordinates stay same, y coordinate  changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.

A(0,0,0)

Reflecting

A(0,0,0)

B(0,5,0)

Reflecting

B(0,-5,0)

C(3,5,0)

Reflecting

C(3,-5,0)

D(3,0,0)

Reflecting

D(3,0,0)

b.)

A(0,0,0)

Moving

A(-2,-3,1)

B(0,-5,0)

Moving

B(-2,-8,1)

C(3,-5,0)

Moving

C(1,-8,1)

D(3,0,0)

Moving

D(1,-3,1)

Hope this helps plz mark brainliest

Step-by-step explanation:

3 0
2 years ago
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AfilCa [17]
1. t = 7
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2 years ago
X-3y=8 what is y in this question
gogolik [260]

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Which data set has a median of 15?
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Step-by-step explanation: Hope this helps ^^

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Identify the center and the radius of a circle that has a diameter with endpoints at (−5, 9) and (3, 5)
marusya05 [52]

Check the picture below, so the circle looks more or less like that one.

well, the center of it is simply the Midpoint of those two points, and its radius is simply half-the-distance between them.

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 -5}{2}~~~ ,~~~ \cfrac{ 5 + 9}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{14}{2} \right)\implies \stackrel{center}{(-1~~,~~7)} \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[3 - (-5)]^2 + [5 - 9]^2}\implies d=\sqrt{(3+5)^2+(-4)^2} \\\\\\ d=\sqrt{8^2+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}~\hfill \stackrel{\textit{half the diameter}}{\cfrac{4\sqrt{5}}{2}\implies \underset{radius}{2\sqrt{5}}}

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