Help i am stuck 48/r = 12

2 answers:

5 0

R=4 It would be this because you have to solve the rational equation by combining expressions and isolating the variable r.

DiKsa [7]2 years ago

4 0

Answer:48/12 is how you solve this so your answer is 4 BRAINLY PLZZZZZZZZZZ

Step-by-step explanation:

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Connor is going to see a movie and is taking his 4 kids. Each movie ticket costs $13 and there are an assortment of snacks avail

EleoNora [17]

If Connor buys 3 snacks at 3 dollars each for himself and his 3 children, the net cost (including movie tickets) would be 88 dollars.

If he buys x amount of snacks, his cost will be determined by C=52+3x or C=52+3*x (both are equivalent, but not everyone knows every way to represent multiplication)

Step-by-step explanation:

To find the cost of the snacks:

Cost = Price*NumberOfSnacks, or C=p*n, or C=pn

To find the unit cost of the snack, find the price, the p value, which is 3 dollars.

Then, find the total number of snacks that will be bought. Since each person receives an equal number of snacks, 3 per person, and there are 4 people (himself and his 3 children), we can multiply 3*4 to find the n value, which is now known to be 12.

C=3*12=36. Add to the 13 dollar per person charge for tickets (52 dollars for 4 people), and you get 88 dollars.

To find the cost if he buys x amount of snacks for himself and his 3 children (assumed to be "everyone"), all costs but the n value are to be the same, creating this equation:

C=52+3x

In this context, 52 is the cost of movie tickets, 3 is the cost of the snacks, and x is the total number of snacks bought. C represents total cost.

5 0

2 years ago

(cot^2x - 1)/(csc^2x) = cos2x

Alex787 [66]

Answer:

Step-by-step explanation:

The idea here is to get the left side simplified down so it is the same as the right side. Consequently, there are 3 identities for cos(2x):

$cos(2x)=cos^2x-sin^2x$ ,

$cos(2x)=1-2sin^2x$ , and

$cos(2x)=2cos^2x-1$

We begin by rewriting the left side in terms of sin and cos, since all the identities deal with sines and cosines and no cotangents or cosecants. Rewriting gives you:

$\frac{\frac{cos^2x}{sin^2x} -\frac{sin^2x}{sin^2x} }{\frac{1}{sin^2x} }$