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guapka [62]
3 years ago
6

How do I solve this ? Help ASAP!!! Steps please

Mathematics
2 answers:
mixer [17]3 years ago
8 0

Answer:

1x + 4

Step-by-step explanation:

add like terms

6x - 3x - 2x = 1x or x then you are left with 4

elena-14-01-66 [18.8K]3 years ago
3 0

Answer:

Step-by-step explanation:

start by gathering all the x together.

6x-3x-2x=1x

what you have left is 1x+4=?x+?

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It is 1 because if you divided 7over 7 it 1 so x=1
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HOW MANY SQUARE NUMBERS IN A PACK OF 52 CARDS
Vesnalui [34]
In a pack of 52 cards, there are 3 square numbers - 1, 4, and 9.

1 × 1 = 1
2 × 2 = 4
3 × 3 = 9
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What is a solution for 8>2x+4
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Read 2 more answers
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

8 0
3 years ago
Nathaniel measured the height of one shelf on his bookcase. He wants to know which books will fit on the shelf. The shelf is 22.
UkoKoshka [18]
PART A - So the ruler is 22.86 centimeters tall, and after measuring with an inch ruler, is also 9 inches tall.
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\frac{22.86cm}{9in} = \frac{x}{1in} \\

Part B - 
Cross multiplying,
(9in)x = (22.86cm)1 \\ &#10;x =  \frac{22.86cm}{9in}= 2.54cm/in

Part C - We can use something called dimensional analysis - multiplying 12 cm by (1in/2.54cm) - in order to change the units from inches to cm. This is possible because 1in = 2.54cm, the fraction 1in/2.54cm = 1. And also the units cm will cancel out.
\frac{12cm}{1} * \frac{1in}{2.54cm} = 4.72in

6 0
3 years ago
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