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kozerog [31]
3 years ago
6

What are the domain and Range of f(x)=Log(x+6)-4

Mathematics
1 answer:
Dominik [7]3 years ago
5 0
f(x)=\log(x+6)-4\\\\\text{The domain:}\\x+6 \ \textgreater \  0\to x \ \textgreater \  -6\to x\in(6;\ \infty)\\\\\text{the range of}\ y=\log x\ \text{is}\ \mathbb{R}\\\text{and the range of}\ y=\log(x+6)\ \text{is}\ \mathbb{R}\\\text{and the range of}\ f(x)=\log(x+6)-4\ \text{is}\ \mathbb{R}

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Find the slope of the line containing the pair of points.<br> ​(6​,5​) and ​(​0,​0)
GarryVolchara [31]

Answer: 5/6

Step-by-step explanation:

Use slope formula of y2-y1/x2-x1. So here it doesn’t matter which is which so I did (6,5) x1 and y1 and (0,0) x2 and y2

So plugging it in it would be 0-5/0-6 = -5/-6 which simplifies to 5/6

7 0
1 year ago
What is 3/10 and 1/5 by using common
astra-53 [7]

Answer:

10

Step-by-step explanation:

5*2=10

10*1=10

so 1/5=2/10

3 0
3 years ago
What is 6/13 rounded to the nearest half?
prohojiy [21]
6/13, when turned into a decimal, is equal to 0.4615384615... And so on. That decimal can be rounded to the nearest hundredth to become 0.46.

Is 0.46 closer to 0.5, or 0? When put on a number line, we can see how close it is to 0.5, which is the same as 1/2.

6/13, rounded to the nearest half, is 1/2.
Hope this helped =)
6 0
3 years ago
Read 2 more answers
Complete the equation to show that 4.2÷0.35 is equal to 420÷35.
Alex

Answer:

12=12

Step-by-step explanation:

4.2÷0.35=420÷35

12=12

12-12=0

0=0

4 0
3 years ago
Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.
Mama L [17]

Multiplying both sides by y^2 gives

xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1

so that substituting v=y^3 and hence \frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx} gives the linear ODE,

\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1

Now multiply both sides by 3x^2 to get

x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2

so that the left side condenses into the derivative of a product.

\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2

Integrate both sides, then solve for v, then for y:

x^3v=\displaystyle\int3x^2\,\mathrm dx

x^3v=x^3+C

v=1+\dfrac C{x^3}

y^3=1+\dfrac C{x^3}

\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}

6 0
3 years ago
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