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3 years ago

2(11x-6)-18 is equal to 22x-30

Mathematics

2 answers:

rewona [7]3 years ago

7 0

Answer: Yes they are equal

Step-by-step explanation: First you have to solve 2(11x-6)-18:

2(11x-6)-18

22x - 12 - 18

22x - 30

They equal!

Vinil7 [7]3 years ago

5 0

Answer:

THE RIGHT HAND SIDE IS EQUAL TO THE LEFT HAND SIDE.

Step-by-step explanation:

2(11x-6)-18 is equal to 22x-30

the above question can as well be expressed as

2(11x-6) - 18 = 22x -30

the next thing is to apply the BODMAS(BRACKET, OF, DIVISION, MULTIPLICATION, ADDITION , SUBTRACTION.) rule by first opening the bracket

22x - 12 -18 = 22x -30

We will proceed to collect the like terms

22x -22x = -30+12+18

0 = -30 +30

0=0

the answer is = 0

THEREFORE THE RIGHT HAND SIDE IS EQUAL TO THE LEFT HAND SIDE.

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Butoxors [25]

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3 years ago

2x-y-3z=7, 3x+y+=10, 5x+y+2z=15

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Answer:

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Step-by-step explanation:

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3 years ago

You get tired of the sand and head up to the amusement park. You can purchase 20 ride tickets for $14 or you can purchase 30 rid

Helga [31]

Answer:

The one with the better deal would be 30 ride tickets for $22.50 this is because you pay less money for more rides.

Step-by-step explanation:

First you divide 20 by 14. Doing this will give you the cost of a ride per ticket.

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4 years ago

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adoni [48]

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Step-by-step explanation:

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4 years ago

Read 2 more answers

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago