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Anarel [89]
3 years ago
14

There are 40 grams of sugar in a 12 oz can of Dr Pepper. If your max intake of sugar permitted for the day is 25 grams, how much

of the Dr. Pepper can you drink?

Mathematics
2 answers:
Zarrin [17]3 years ago
6 0

To answer this question, we can set up a proportion:

\frac{40g}{12oz} =\frac{25g}{x}

Then we cross multiply and solve for x:

40x=300

x=7.5oz

So you can drink 7.5 oz of the Dr. Pepper in one day.

Thepotemich [5.8K]3 years ago
6 0

Answer:

7.5 oz can of Dr.Pepper.

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Algebraic Equations.

Since we are given 3 specific values as well as a variable, and because this is a ratio problem we can use the <u><em>Rule of Three technique</em></u> in order to calculate the amount of Dr.Pepper that you can drink. This Technique is demonstrated in the picture attached below.

Let's apply this technique with the values given to us in the question.

40 grams   ⇒   12 oz

25 grams   ⇒   x

\frac{25*12}{40} = x

7.5oz = x

Therefore, we can see that with a limit of 25 grams you can drink a maximum of a 7.5 oz can of Dr.Pepper.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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postnew [5]

Answer:

Step-by-step explanation:

The surface area of a sphere with radius r is given by

S = 4 pi r^2

for a radius of 11 units,

S =  4 pi 11^2 = 484 pi = 1520.5 sq. units

This does not correspond to any of the answers.

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3 years ago
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A circle has an area of 78.5 square inches what is the radius of the circle
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Answer:

Step-by-step explanation:

Area of circle = 78.5 in²

πr² = 78.5

3.14*r² = 78.5

r^{2}=\frac{78.5}{3.14}\\\\=\frac{78.5*100}{3.14*100}\\\\=\frac{7850}{314}\\\\r^{2}=25\\r=\sqrt{25}=\sqrt{5*5}\\\\

r =  5 in

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3 years ago
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Will mark brainliest for the correct answer!
romanna [79]

Part (a)

Focus on triangle PSQ. We have

angle P = 52

side PQ = 6.8

side SQ = 5.4

Use of the law of sines to determine angle S

sin(S)/PQ = sin(P)/SQ

sin(S)/(6.8) = sin(52)/(5.4)

sin(S) = 6.8*sin(52)/(5.4)

sin(S) = 0.99230983787513

S = arcsin(0.99230983787513)

S = 82.889762826274

Which is approximate

------------

Use this to find angle Q. Again we're only focusing on triangle PSQ.

P+S+Q = 180

Q = 180-P-S

Q = 180-52-82.889762826274

Q = 45.110237173726

Which is also approximate.

A more specific name for this angle is angle PQS, which will be useful later in part (b).

------------

Now find the area of triangle PSQ

area of triangle = 0.5*(side1)*(side2)*sin(included angle)

area of triangle PSQ = 0.5*(PQ)*(SQ)*sin(angle Q)

area of triangle PSQ = 0.5*(6.8)*(5.4)*sin(45.110237173726)

area of triangle PSQ = 13.0074347717966

------------

Next we'll use the fact that RS:SP is 2:1.

This means RS is twice as long as SP. Consequently, this means the area of triangle RSQ is twice that of the area of triangle PSQ. It might help to rotate the diagram so that line PSR is horizontal and Q is above this horizontal line.

We found

area of triangle PSQ = 13.0074347717966

So,

area of triangle RSQ = 2*(area of triangle PSQ)

area of triangle RSQ = 2*13.0074347717966

area of triangle RSQ = 26.0148695435932

------------

We're onto the last step. Add up the smaller triangular areas we found

area of triangle PQR = (area of triangle PSQ)+(area of triangle RSQ)

area of triangle PQR = (13.0074347717966)+(26.0148695435932)

area of triangle PQR = 39.0223043153899

------------

<h3>Answer: 39.0223043153899</h3>

This value is approximate. Round however you need to.

===========================================

Part (b)

Focus on triangle PSQ. Let's find the length of PS.

We'll use the value of angle Q to determine this length.

We'll use the law of sines

sin(Q)/(PS) = sin(P)/(SQ)

sin(45.110237173726)/(PS) = sin(52)/(5.4)

5.4*sin(45.110237173726) = PS*sin(52)

PS = 5.4*sin(45.110237173726)/sin(52)

PS = 4.8549034284642

Because RS is twice as long as PS, we know that

RS = 2*PS = 2*4.8549034284642 = 9.7098068569284

So,

PR = RS+PS

PR = 9.7098068569284 + 4.8549034284642

PR = 14.5647102853927

-------------

Next we use the law of cosines to find RQ

Focus on triangle PQR

c^2 = a^2 + b^2 - 2ab*cos(C)

(RQ)^2 = (PR)^2 + (PQ)^2 - 2(PR)*(PQ)*cos(P)

(RQ)^2 = (14.5647102853927)^2 + (6.8)^2 - 2(14.5647102853927)*(6.8)*cos(52)

(RQ)^2 = 136.420523798282

RQ = sqrt(136.420523798282)

RQ = 11.6799196828694

--------------

We'll use the law of sines to find angle R of triangle PQR

sin(R)/PQ = sin(P)/RQ

sin(R)/6.8 = sin(52)/11.6799196828694

sin(R) = 6.8*sin(52)/11.6799196828694

sin(R) = 0.4587765387107

R = arcsin(0.4587765387107)

R = 27.3081879220073

--------------

This leads to

P+Q+R = 180

Q = 180-P-R

Q = 180-52-27.3081879220073

Q = 100.691812077992

This is the measure of angle PQR

subtract off angle PQS found back in part (a)

angle SQR = (anglePQR) - (anglePQS)

angle SQR = (100.691812077992) - (45.110237173726)

angle SQR = 55.581574904266

--------------

<h3>Answer: 55.581574904266</h3>

This value is approximate. Round however you need to.

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