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3 years ago

Find the length of the hypotenuse of a right angle if a = 6 and b = 10

Mathematics

2 answers:

defon3 years ago

8 0

Answer:

11.66

Step-by-step explanation:

c² = a² + b²

c²= 6² + 10²

c²= 136

c= √136

c= 11.66 ( to two decimals)

Anton [14]3 years ago

8 0

Answer:

2sqrt34

Step-by-step explanation:

6^2+10^=136

sqrt136= 2sqrt34

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In order to solve this sentence you must make terms out of the words. Answer: 1 + (1x)

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3 years ago

Please Simplify the equation (-6) ^-2

xeze [42]

Answer:

1/36

Step-by-step explanation:

(-6) ^-2

The negative in the exponent means in goes from the numerator to the denominator

(-6) ^-2 = 1/ (-6)^2

We know that

(-6)^2 = -6*-6 = 36

(-6) ^-2 = 1/ 36

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3 years ago

Read 2 more answers

A car rental company charges a one-time application fee of 20 dollars, 45 dollars per day, and 15 cents per mile for its cars. W

crimeas [40]

Answer:

$C(d,m) = 20 + 45d + 0.15m$

Step-by-step explanation:

In this problem, we have that the cost of renting a car C is a function of the number of days d, and the number of miles driven, m.

There is also a fixed cost F.

So the equation for the cost of renting a car is:

$C(d,m) = F + a*d + b*m$

In which a is the daily cost and b is the cost per mile.

A car rental company charges a one-time application fee of 20 dollars, 45 dollars per day, and 15 cents per mile for its cars.

This means that $F = 20, a = 45, b = 0.15$

Write a formula for the cost, C, of renting a car as a function of the number of days, d, and the number of miles driven, m.

$C(d,m) = 20 + 45d + 0.15m$

6 0

3 years ago

A drag racer has two parachutes, a main and a backup, that are designed to bring the vehicle to a stop after the end of a run. S

sweet-ann [11.9K]

Answer:

the probability that one parachute deploys is P(one parachute deploys)= 0.9988 (99.88%)

Step-by-step explanation:

for event A= first parachute deploys , then P(A)=0.97

for event B= second parachute deploys, then

P(C∩B)= P(B/C)*P(C)

where

event C= first parachute fails

P(C) = 1-P(A)= 1-0.97 =0.03

P(B/C) = probability that second parachute deploys, when the fist one fails =0.96

P(C∩B)= probability that fist parachute fails and second parachute deploys

then

P(C∩B)= P(B/C)*P(C)=0.03*0.96 = 0.0288

then

P(one parachute deploys)= P(A) + P(C∩B) - P(A∩(C∩B)) ,

but since A and C are mutually exclusive events P(A∩(C∩B)) =0

thus

P(one parachute deploys)= P(A) + P(C∩B) =0.97 + 0.0288 = 0.9988

P(one parachute deploys)= 0.9988

6 0

3 years ago

(TCO 3) How many different teams can be organized consisting of two men, two women, and two children? We have the following grou

Nady [450]

Answer:

900 teams

Step-by-step explanation:

The order in which the people are selected is not important(John and Josh is the same team as Josh and John), which means that the combinations formula is used 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:

Two men(from a set of 6).

Two women(from a set of 5).

Two children(from a set of 4).

How many different teams can be organized consisting of two men, two women, and two children?

$T = C_{6,2}*C_{5,2}*C_{4,2} = \frac{6!}{2!4!}*\frac{5!}{2!3!}*\frac{4!}{2!2!} = 15*10*6 = 900$

900 teams

4 0

3 years ago