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3 years ago

A skateboarding ramp is 12 in high and rises at an angle of 17 degrees. How long is the base of the ramp?

Mathematics

2 answers:

tia_tia [17]3 years ago

7 0

Answer:

39.3 inches

Step-by-step explanation:

Let x represent the length of base of ramp.

We have been given that a skateboarding ramp is 12 in high and rises at an angle of 17 degrees. We are asked to find the base of the skateboard.

The height of ramp, the base of ramp and angle of elevation make a right triangle with ground, where, 12 inch is opposite side and base of ramp is adjacent side of angle of elevation.

$\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}$

$\text{tan}(17^{\circ})=\frac{12}{x}$

$x=\frac{12}{\text{tan}(17^{\circ})}$

$x=\frac{12}{0.305730681459}$

$x=39.25023$

$x\approx 39.3$

Therefore, the base of ramp is 39.3 inches long.

antoniya [11.8K]3 years ago

4 0

Check the picture below.

make sure your calculator is in Degree mode.

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3 years ago

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ANTONII [103]

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3 years ago

The square below has an area of 16 +82 + square meters.

svlad2 [7]

Answer:

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Step-by-step explanation:

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3 years ago

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Choose 5 cards from a full deck of 52 cards with 13values (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and 4 kinds(spade, diamond, h

Delvig [45]

Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

$T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182$

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

$T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148$

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

$T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378$

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

$T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899$

So 2899 possible ways.

4 0

3 years ago