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2 years ago

What is the value of x in the equation 3x = 4 + x + 6?

Mathematics

2 answers:

Ket [755]2 years ago

6 0

Step-by-step explanation:

x = 5 is the answer for the question

Nastasia [14]2 years ago

5 0

Answer:

x = 5

Step-by-step explanation:

3x = 4 + x + 6

2x = 10

x = 5

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A x + b y is equal to 12

Nikitich [7]

Answer:

y= -a*x + 12

b b

Step-by-step explanation:

ax+by=12 (subtract "ax" from both sides so that the one on the left will become zero and we will have "by" )

by= -ax+12(divide both side by "by" so that we will have the equation as "y=mx+b")

finally the result will be:

y= -a*x + 12

b b

7 0

2 years ago

Write an equation to represent the relationship found in the table

PilotLPTM [1.2K]

Answer:

15= 5x+ 2

Step-by-step explanation:

y= mx+b

y is the intercept,

8 0

2 years ago

How do i find the ratio 2:9

Ganezh [65]

The ratio of 2:9 has to be 2:9 do to the fact you can't simplify the fraction 2/9.
Hoped I helped.

6 0

3 years ago

If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(

timama [110]

Use the change-of-basis identity,

$\log_x(y) = \dfrac{\ln(y)}{\ln(x)}$

to write

$xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}$

Use the product-to-sum identity,

$\log_x(yz) = \log_x(y) + \log_x(z)$

to write

$xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}$

Redistribute the factors on the left side as

$xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}$

and simplify to

$xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)$

Now expand the right side:

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}$

Simplify and rewrite using the logarithm properties mentioned earlier.

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1$