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Alchen [17]
4 years ago
13

3. A sum of $2700 is to be given in the form of 63 prizes. If the prize is of either $100 or $25, find the number of prizes of e

ach type.
Mathematics
1 answer:
34kurt4 years ago
8 0
There has to be a total of 63 boxes each with a prize in them.

There are two types of prizes: The first type is $25 and second type is $100. 

If you make all the prizes $100, you will only end up with 27 prizes. ($2700 / 100), which is too little. 

If you make all the prizes $25, you will only end up with 108 prizes ($2700/ 25), which is too way many. 

The answer has to be combination of prizes More $25 prizes will have an outcome of more prizes in total. 
More $100 prizes will have an outcome of less prizes in total. 

The answer would come out to

48 prizes of $25 each ($1200) 
15 prizes of $100 each ($1500)

which adds to the full amount of $2700 with 63 prizes.
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Please help me with this problem​
solniwko [45]

Answer:

  see attached

Step-by-step explanation:

The area of a square is the square of the side length.

For example, (√125)² = 125.

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The side length of a square is the square root of the area.

For example, √98 = √(7²·2) = 7√2.

5 0
3 years ago
You buy a movie for 19.99 and a set of earphones for 12.49 how much is the bill before taxes
-BARSIC- [3]
19.99+12.49=32.48
19.99 for the movie and 12.49 for the earphones together they make 32.48
7 0
3 years ago
Read 2 more answers
Jasmine drew a rectangle with an area of 4 square inches. Jasmine drew a new scaled version of the original rectangle using a sc
xz_007 [3.2K]

Answer:

the answer is 10 im pretty sure

Step-by-step explanation:

6 0
3 years ago
If sin theta = 8/17 and cot theta < 0, what is sec theta?
klasskru [66]

Answer:

-\frac{17}{15}

Step-by-step explanation:

By definition, \cot \theta=\frac{1}{\tan \theta} and \sec \theta=\frac{1}{\cos \theta}. Since since \cot \theta is negative, \tan \theta must also be negative, and since \sin \theta is positive, we must be in Quadrant II.

In a right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. The cosine of an angle in a right triangle is equal to its adjacent side divided by the hypotenuse. Therefore, we can draw a right triangle in Quadrant II, where the opposite side to angle theta is 8 and the hypotenuse of the triangle is 17.

To find the remaining leg, use to the Pythagorean Theorem, where a^2+b^2=c^2, where c is the hypotenuse, or longest side, of the right triangle and a and b are the two legs of the right triangle.

Solving, we get:

8^2+b^2=17^2,\\b^2=17^2-8^2,\\b^2=\sqrt{17^2-8^2}=\sqrt{225}=15

Since all values of cosine theta are negative in Quadrant II, all values of secant theta must also be negative in Quadrant II.

Thus, we have:

\sec\theta=\frac{1}{\cos \theta}=-\frac{1}{\frac{15}{17}}=\boxed{-\frac{17}{15}}

6 0
3 years ago
2.
Valentin [98]

Answer:

et55

Step-by-step explanation:

3 0
3 years ago
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