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3 years ago

HELP PLEASE!! . . . . .

Mathematics

2 answers:

max2010maxim [7]3 years ago

7 0

Answer:

Step-by-step explanation:

because they spent the least of money on flowers

ANTONII [103]3 years ago

7 0

Answer:

They spent more on signs than flower boxes is the correct answer.

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What is percent is 14 out of 25?

natka813 [3]

Answer: 3.50

Step-by-step explanation:

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3 years ago

A full glass of water can hold 1/6 of a bottle.

uysha [10]

Answer:

21 glasses of water.

Step-by-step explanation:

You can do a proportion.

$\fraction{1 glass}{1/6 of a bottle)$ or

1 glass/1/6 of a bottle.

x glasses/3.5 bottles.

You can cross multiply.

1/6 times x is 1/6x

1 times 3.5 is 3.5/

1/6x=3.5.

do the reciprocal of 1/6 which is 6.

So, 6 times 3.5 is 21.

Therefore, you have 21 glasses of water.

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4 years ago

One day a rickshaw earned $80 out of his earnings he spent $68/5on tea and $51/2 on food and $22/5 on repairs of the rickshaw ho

Tamiku [17]

Use this app called Slader it should help

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3 years ago

A carton in the shape of a cube measuring 8 in. tall is filled with colored plastic cubes measuring 1 in.across each face. The c

DerKrebs [107]

You can fir 8x8x8 cubes in the carton, that means you can fit 512 cubes in the carton.
the total cost would be, 0.25 x 512 = $128

7 0

3 years ago

Read 2 more answers

I don’t know how to solve this

Lady_Fox [76]

Answer:

$\theta =2\pi k,\ \ k\in Z\ \\\text{or}\ \\\theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

Step-by-step explanation:

Given:

$\cos \theta-\sqrt{3}\sin \theta=1$

Divide this equation by 2:

$\dfrac{1}{2}\cos \theta-\dfrac{\sqrt{3}}{2}\sin \theta=\dfrac{1}{2}$

Note that

$\cos \dfrac{\pi }{3}=\dfrac{1}{2}\\ \\\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$

So, the previous equation is

$\cos \dfrac{\pi}{3}\cdot \cos \theta-\sin \dfrac{\pi}{3}\cdot \sin \theta=\dfrac{1}{2}$

Remind that

$\cos x\cos y-\sin x\sin y=\cos (x+y),$

then

$\cos \left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{2}$

The solution of this equation is

$\dfrac{\pi}{3}+\theta=\pm \arccos \dfrac{1}{2}+2\pi k,\ \ k\in Z\\ \\\dfrac{\pi}{3}+\theta=\pm \dfrac{\pi}{3}+2\pi k,\ \ k\in Z\\ \\\theta =2\pi k,\ \ k\in Z\ \text{or}\ \theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

8 0

3 years ago