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3 years ago

best answer will be awarded the brainliest and 5 stars please evaluate 6/2 and remember to show your work

1 answer:

8 0

6/2 is the number 6 being divided in half by 2; to figure this out just count your 2 table factors! Shown work:

2 x 2 = 4
2 x 3 = 6!

So 6 divided by 2 (6/2) is just the oppoposite of 2 x 3 = 6

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What is the volume of the figure below, which is composed of two cubes with side lengths of 6 units?

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Which equation is equivalent to..

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Step-by-step explanation:

$1/3)^{x}$ = $\frac{1}{3^{x} }$ = $3^{-x}$

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3 years ago

Please someone help me to prove this..

Pachacha [2.7K]

Answer: see proof below

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

Use the following Sum to Product Identities:

$\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)$

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}$

$\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}$

$\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}$ $\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}$

$\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}$

$\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}$

$\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10$

LHS = RHS: cot 10 = cot 10 $\checkmark$

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3 years ago

The table shows temperatures below freezing measured in different units. Complete the equation in standard form to represent the

dezoksy [38]

The relationship between the temperature in celsius to the temperature in Fahrenheit is C = (5/9)(F - 32)

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Let F represent temperature measured in degrees Fahrenheit, and C, a temperature measured in degrees Celsius.

At 0°C, the temperature is 32°F. Also at 100°C, the temperature is 212°F, hence:

$C-0=\frac{100-0}{212-32} (F-32)\\\\C=\frac{5}{9} (F-32)$

The relationship between the temperature in celsius to the temperature in Fahrenheit is C = (5/9)(F - 32)

Find out more on equation at: brainly.com/question/297283

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3 years ago