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3 years ago

11

Wrire a word problem for 3/5 times 2/3.

1 answer:

goldfiish [28.3K]3 years ago

7 0

Reduce the numbers with greatest common divisor 3

1/5 times 2

Calculate the product

2/5

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Please help due today NO LINKS WILL REPORT!!

katen-ka-za [31]

Answer:

{0,1,2,3}

because whole numbers always starts from zero...

6 0

2 years ago

Three friends grow groups of sunflowers. Some think that Anna’s flowers are the tallest because the mean height of her flowers i

algol [13]

We have to find the graph where the mean of the flower heights is higher and the median is lower.

1st graph
M = ( 3 + 3 + 3 + 3 + 3 ) / 7
M = 15 / 7
Medium = 3
15 / 7 < 3 ( False )
2st graph
M = ( 2 + 6 + 2 + 1 + 1 + 3 ) / 6
M = 15 / 6
Medium = 4 / 2 = 2
15 / 6 > 2
So this is the requested chart.
Hope this helps, good studies.

7 0

2 years ago

Looking at Morningstar’s Financial Health ratios, what has happened to HP’s liquidity position over the past 10 years?

solong [7]

Answer:

Step-by-step explanation:

Huh lol ok

8 0

3 years ago

PLZ SOMEONE HELP ME

Sphinxa [80]

Answer:

9 teaspoons of cocoa

Step-by-step explanation:

3*3=9

6 0

2 years ago

In Exercises 5-7, find all the exact t-values for which the given statement is true,

bearhunter [10]

Answer: See Below

<u>Step-by-step explanation:</u>

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians: t = 0π + 2πn

In degrees: t = 0° + 360n

******************************************************************************

$6)\quad sin (t) = \dfrac{1}{2}$

Where on the Unit Circle does $sin = \dfrac{1}{2}$

<em>Hint: sin is only positive in Quadrants I and II</em>

$\text{Answer: at}\ \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n$

In degrees: t = 30° + 360n and 150° + 360n

******************************************************************************

$7)\quad tan (t) = -\sqrt3$

Where on the Unit Circle does $\dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)$

<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>

$\text{Answer: at}\ \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n$

In degrees: t = 120° + 360n and 300° + 360n

4 0

2 years ago