All categories

3 years ago

Solve for the missing side in the following right triangle: hypotenuse = 17, leg = 8, leg = ?.

Mathematics

1 answer:

krek1111 [17]3 years ago

7 0

Hyp^2 = leg1^2 + leg2^2
hyp^2 -leg1^2 = leg2^2
17^2 - 8^2 = leg2^2
leg2^2 = 289 -64
leg2^2 = 225
leg2 = 15

You might be interested in

N x n² da como resultado=

kykrilka [37]

N x n2 da como resultado

3 0

3 years ago

Read 2 more answers

Jon recently drove to visit his parents who live 280 miles away. On his way there his average speed was 9 miles per hour faster

Luba_88 [7]

Answer:

11 mph and 20 mph

Step-by-step explanation:

Represent his average speed going by r1 and his average speed returning by r2. We know that r1 = r2 + 9.

Recall that distance = rate times time, so time = distance / rate.

Time spent going was (280 mi) / r1, or (280 mi) / (r2 + 9 mph).

Time spend returning was (280 mi) / r2.

The total time was 14 hrs, so (280 mi) / (r2 + 9 mph) + (280 mi) / r2 = 14 hrs

Note that there is only one variable here: r2. Find r2, and then from r2, find r1:

Dividing all 3 terms by 14 hrs yields:

20 20

---------- + ----------- = 1

r2 + 9 r2

The LCD here is r2(r2 + 9). Thus, we have:

20r2 (r2 + 9)(r2)

------------------- = 1 or ------------------

(r2 + 9)(r2) (r2 + 9)(r2)

Then 20(r2) = (r2)^2 + 9(r2). This is reducible by dividing all terms by r2:

20 = r2 + 9, or 11 = r2. Then r1 = 11 + 9, or 20.

The two rates were 11 mph and 20 mph.

8 0

3 years ago

Pleaseee help me with this problem

Liula [17]

Answer:

Step-by-step explanation:

4 0

3 years ago

You throw a ball up and its height h can be tracked using the equation h=2x^2-12x+20.

postnew [5]

This problem seems to be wrong because no minimum point was found and no point of landing exists

Answer:

1) There is no maximum height

2) The ball will never land

Step-by-step explanation:

Derivatives

Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.

Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that

If f''(a) is positive, then x=a is a minimum

If f''(a) is negative, then x=a is a maximum

The function provided in the question is

$h(x)=2x^2-12x+20$

Let's find the first derivative

$h'(x)=4x-12$

solving h'=0:

$4x-12=0$

x=3

Computing h''

h''(x)=4

It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height

To find when will the ball land, we set h=0

$2x^2-12x+20=0$