All categories

2 years ago

Oki so this is the 3rd time ive asked this and i keep forgetting to put in the qeustion im very sorry

Mathematics

2 answers:

Zarrin [17]2 years ago

3 0

The answer is greater than one. :)

Vadim26 [7]2 years ago

3 0

7/6 so it's greater than haha

You might be interested in

Someone bought three t-shirts for $15 and nine $0. 90 each including tax how much did he spend

Goshia [24]

You multiply three( # of shirts) and 15.90 ( price)

The answer is $47.70 is how much he spend

6 0

3 years ago

A triangle has two sides of length 20 and 11. What is the largest possible whole-number length for the third side?

charle [14.2K]

Answer:

Step-by-step explanation:

20+11=31

It can't be exactly 31 but it can be any number less than 31, so the nearest whole number down is 30

8 0

2 years ago

Three black beads, two red beads, and a yellow bead are arranged in a line. What is the probability that no black bead is adjace

shtirl [24]

Answer:

Probability of no black bead is adjacent to the yellow bead is = 0.966

Step-by-step explanation:

First to arrange all the beads without restrictions = 6! = 720

Probability the black bead is adjacent to Yellow bead = 4!/6! = 24 /720

= 0.0333

Probability of no black bead is adjacent to the yellow bead is

= 1- Probability the black bead is adjacent to Yellow bead

= 1-0.033

= 0.966

6 0

3 years ago

Jana ran the first 3 and a half miles of a 5 mile race in 1 3rd of an hour. what was her average rate, in mph, for the first par

Lisa [10]

Answer:

always try to draw a diagram first

Step-by-step explanation:

Jana ran The first 3.5 miles in 1/3 of an hour. remember the problem tells us that we want our answer in miles per hour or mph. so we have to set up our problem so that our units match in the end. The diagram above shows 3.5 mi over 1/3 hour. so we must divide 3.5 miles by 1/3 hour

note:

$3 \frac{1}{2} mi = \frac{7mi}{2}$

therefore

$\frac{7mi}{2} \: \div \: \frac{1hr}{3} =$

$\frac{7mi}{2} \times \frac{3}{1hr} = \frac{21mi}{2hr} = 10 \frac{1}{2} \frac{mi}{hr}$

7 0

2 years ago

Can someone explain to me why are we adding 2kpi when we are doing zeros for sin and cos, but adding kpi when doing zeros for tg

makvit [3.9K]

on the first exercise, you got a solution angle of π/18, that's a good solution for the I Quadrant only, however, on a circle, we have angles that go from 0 to 2π, however we can always keep on going around and continute to 2π + π/2 or 3π or 4π, or 115π/3 or 1,000,000π/18 and so on, and we're really just going around the circle many times over, getting a larger and larger angle, same circular motion.

π/18 on that exercise works for the I Quadrant, however if we continue and go around say 2π, we'll find that 2π/3 + π/18 is a coterminal angle with π/18, and thus that angle has also the same sine value.

π/18 + 2kπ/3 , where k = integer, is a way to say, all angles around the circle that look like this have the same sine, namely

π/18 + 2(1)π/3

π/18 + 2(2)π/3

π/18 + 2(3)π/3

π/18 + 2(5)π/3

π/18 + 2(99999999)π/3

.....

so using the "k" as some sequence multiplier, is a generic notational way to say, "all these angles".

you'll also find that "n" is used as well for the same notation, say for example

2π/3 + 2πn.

8 0

2 years ago