Answer:
<h3>
♁ Question : Solve for x</h3>
<h3>♁ Step - by - step explanation</h3>
Move 12x to L.H.S ( Left Hand Side ) and change it's sign
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Move 7 to R.H.S ( Right Hand Side) and change it's sign
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Subtract 12x from 15x
Remember that only coefficients of like terms can be added or subtracted.
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Add the numbers : 2 and 7
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Divide both sides by 3
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The value of x is
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☄ Now, let's check whether the value of x is 3 or not!
<h3>
☥ Verification :</h3>
L.H.S = R.H.S ( Hence , the value of x is 3 ).
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<h3>✒ Rules for solving an equation :</h3>
- If an equation contains fractions ,multiply each term by the L.C.M of denominators.
- Remove the brackets , if any.
- Collect the terms with the variable to the left hand side and constant terms to the right hand side by changing their sign ' + ' into ' - ' and ' - ' into ' + ' .
- Simplify and get the single term on each side.
- Divide each side by the coefficient of variable and then get the value of variable.
Hope I helped!
Have a wonderful time ! ツ
~TheAnimeGirl
Answer:
x = 1/4
Step-by-step explanation:
-7x - (-5-x) = -9(2x-1) -1
-7x + 5 + x = -18x + 9 - 1
combine like terms:
-6x + 5 = -18x + 8
add 18x to each side of the equation:
12x + 5 = 8
subtract 5 from each side:
12x = 3
divide both sides by 12:
x = 3/12 = 1/4
Answer:
a is the ans;8/15
Step-by-step explanation:
we have tanb = p/ b
so,by taking reference angle b we get,
p=8 and b= 15.
therefore; tanb=8/15
In exponential form it is given as, 8⁻³ =
<u>Step-by-step explanation:</u>
We have to write the given equation in the exponential form, that is in the form of base to the power value.
Given:
log₈ ( ) = -3
Here the base is 8.
In the exponential form, it is written as,
base to the power 3 = 512
Here the reciprocal of 512 is given, so,
base to the power -3 =
So the exponential form is given as, 8⁻³ =
Because if you multiply a negative by a negative, it is positive, right? So, in this instance it is cubed, which means that the number is multiplied by itself 3 times, so the positive is multiplied by the negative, which makes it negative