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kkurt [141]
3 years ago
7

Given y = log3(x + 4), what is the domain?

Mathematics
2 answers:
Basile [38]3 years ago
6 0
The logaritm of b base a

log_ab

Conditions for the base a: a>0 \ and \ a\ne 1 
Conditions for the number b: b>0

b=x+4\\\\ x+4>0 \Rightarrow x>-4

Answer
D: \ x\in (-4;\infty) 
dem82 [27]3 years ago
5 0
For a logarithmic function, we have a restriction on the domain.
Since log(0) isn't defined, we say that there is an asymptote at x = 0.
Thus, for the regular logarithmic function y = log(x), x > 0.

We can then say (x + 4) > 0, since that's when the function of a logarithm is defined as.
x + 4 > 0
x > -4

Thus, the domain of the logarithmic function is x > -4, where x is a real integer.
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3 years ago
You have two circles, one with radius r and the other with radius R. You wish for the difference in the areas of these two circl
gtnhenbr [62]

Answer:

maximum difference of radii =(r-R)=\frac{1}{2\pi ^{2}}

Step-by-step explanation:

We know that area of circle is given by

A=\pi \times (radius)^{2}

For circle with radius 'r' we have

A_{1}=\pi \times (r)^{2}

For circle with radius 'R' we have

A_{2}=\pi \times (R)^{2}

Now according to given condition we have

A_{1}-A_{2}\leq \frac{5}{\pi }

\Rightarrow \pi r^{2}-\pi R^{2}\leq \frac{5}{\pi }\\\\\Rightarrow (r^{2}-R^{2})\leq \frac{5}{\pi ^{2}}\\\\(r+R)(r-R)\leq \frac{5}{\pi ^{2}}\\\\\because (a^{2}-b^{2})=(a+b)(a-b)\\\\(r+R)=10(Given)\\\\\Rightarrow(r-R)\leq \frac{5}{10\pi ^{2}}\\\\\therefore (r-R)\leq\frac{1}{2\pi ^{2}}

Thus maximum difference of radii =(r-R)=\frac{1}{2\pi ^{2}}

6 0
4 years ago
The cost of 3 tacos and a juice is $7. The cost of 4 tacos and 2 juices is $10. If t = the cost of a taco and j = the cost of a
marin [14]

Answer:

t=2

j=1

Step-by-step explanation:

3 tacos and a juice is $7

3t+j=7

4 tacos and 2 juices is $10

4t+2j=10

We have a simultaneous equation. It can be solved either by substitution method or Elimination method. For the purpose of this question, the elimination method will be used

3t+j=7 Equation 1

4t+2j=10 Equation 2

To eliminate j, we multiply Equation 1 by 2, the coefficient of j in equation 2, so that in both equations, j can have the same coefficient.

2(3t+j=7)

6t+2j=14 Equation 3

We now have

4t+2j=10 Equation 2

6t+2j=14 Equation 3

We subtract equation 2 from equation 3, to eliminate j

6t-4t=2t

2j-2j=0

14-10=4

We have 2t=4

Divide both sides by 2

2t/2=4/2

t=4/2

t=2

Substitute t for 2 in equation 1, to get j

3t+j=7

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6+j=7

j=7-6

j=1

One taco cost $2 while 1 juice cost $2

3 0
3 years ago
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