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Norma-Jean [14]
3 years ago
12

) Check whether the value given in the bracket is a solution to the given equation or not. x – 7 = 17, (x = 10

Mathematics
1 answer:
S_A_V [24]3 years ago
4 0

Answer:

I think you need to change the subtract sign to addition or you change the 10 to 24

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Write the equation 9y = 12x + 0.2 in standard form. Identify A, B, and C.
uysha [10]

Answer:

A = -12, B = 9 and C = .2

Step-by-step explanation:

So it just looks like we need to use algebra to move everything around to get it in the form Ax + By = C

So, let's start with 9y = 12x + 0.2.  From standard form we have all terms wth variables on one side, so let's do that.  the right sie has both a variable term and non variable term, so let's get rid of the variable term.  The variable term is 12x, so to get rid of it you subtract 12x from both sides.

9y = 12x + 0.2

9y - 12x = 0.2

To make it exactly in standard form just rarrnge.

-12x + 9y = 0.2

So now we have A = -12, B = 9 and C = .2

5 0
3 years ago
What is the solution of <br> x/2=6
Lina20 [59]

Answer:

12

Step-by-step explanation:

12/2=6

replace x with 12

4 0
3 years ago
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What is midpoint 85 90?
monitta
The midpoint of 85 and 90 is 87
8 0
3 years ago
HELP ME ASAP WITH MATH MONEY &amp; WAGES
Dennis_Churaev [7]

Answer:

96,220

Step-by-step explanation:

187,400 x .3 (30%)= 56,220

56220+40000= *96,220

5 0
3 years ago
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Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse
Drupady [299]

Answer:

a'(d) = \frac{d}{5} + \frac{3}{5}

a(a'(d)) = a'(a(d)) = d

Step-by-step explanation:

Given

a(d) = 5d - 3

Solving (a): Write as inverse function

a(d) = 5d - 3

Represent a(d) as y

y = 5d - 3

Swap positions of d and y

d = 5y - 3

Make y the subject

5y = d + 3

y = \frac{d}{5} + \frac{3}{5}

Replace y with a'(d)

a'(d) = \frac{d}{5} + \frac{3}{5}

Prove that a(d) and a'(d) are inverse functions

a'(d) = \frac{d}{5} + \frac{3}{5} and a(d) = 5d - 3

To do this, we prove that:

a(a'(d)) = a'(a(d)) = d

Solving for a(a'(d))

a(a'(d))  = a(\frac{d}{5} + \frac{3}{5})

Substitute \frac{d}{5} + \frac{3}{5} for d in  a(d) = 5d - 3

a(a'(d))  = 5(\frac{d}{5} + \frac{3}{5}) - 3

a(a'(d))  = \frac{5d}{5} + \frac{15}{5} - 3

a(a'(d))  = d + 3 - 3

a(a'(d))  = d

Solving for: a'(a(d))

a'(a(d)) = a'(5d - 3)

Substitute 5d - 3 for d in a'(d) = \frac{d}{5} + \frac{3}{5}

a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}

Add fractions

a'(a(d)) = \frac{5d - 3+3}{5}

a'(a(d)) = \frac{5d}{5}

a'(a(d)) = d

Hence:

a(a'(d)) = a'(a(d)) = d

7 0
2 years ago
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