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3 years ago

What is the solution for 3x =9

Mathematics

2 answers:

mixas84 [53]3 years ago

6 0

x = 3

Divide each term by 3 and simplify.

Vinvika [58]3 years ago

5 0

The answer to your question is 3

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Please help Im confused and I need this to be done ASAP what is the range

il63 [147K]

Answer:

All Real Number greater than 0 or

(0, positive infinity)

Step-by-step explanation:

The range is all possible y values of a function. We can tell a range by looking at the lowest y vaule then going to the highest possible y value.

Look at the graph. There is a open circle at y=0. So this means we are not going to include 0 in our function. The graph approaches f(x) to positive Infinity so our highest y value is Infinity.

So we know that the function y values cannot be zero and it must be positive. So our range is

All Real Number greater than 0. Or

(0, positive infinity)

3 0

2 years ago

Pls help............

klemol [59]

Answer: Your picture will not show up, can you post again?

Step-by-step explanation:

5 0

3 years ago

Can someone explain this to me? I'm not entirely sure this question has enough information to be solved. Thanks!

dedylja [7]

Question 1

$DC=AC-AD=20-16=4$

Question 2

By the geometric mean theorem,

$BD^2 = (4)(16)=64 \\ \\ BD=8$

Question 3

By the Pythagorean theorem,

$AB^2 = 8^2 + 16^2 \\ \\ AB^2 = 320 \\ \\ AB=8\sqrt{5}$

Question 4

By the Pythagorean theorem,

$BC^2 = 8^2 +4^2 \\ \\ BC^2 = 80 \\ \\ BC=4\sqrt{5}$

8 0

1 year ago

F(x)=x/3 -2 and g(x)=2x^2+x-3 find (f+g)(x)

Serjik [45]

Answer:

(f+g)(x) = 2x^2 + 4x/3 -5

Step-by-step explanation:

so, (f+g)(x) be y

y = f(x) + g(x)

y = x/3 -2 + 2x^2 + x - 3 = 2x^2 + 4x/3 -5

6 0

2 years ago

Analyzing a Graph In Exercise, analyze and sketch the graph of the function. Label any relative extrema, points of inflection, a

vivado [14]

Answer:

analyzed & sketched

Step-by-step explanation:

We are given the function: y = In(x + 3)

So first we need to take the first derivative to find the interval of increase and decrease, then the second derivative to find the interval of concave up and concave down.

$y'=\frac{1}{x+3}$

$y''=-\frac{1}{(x+3)^2}$

So it is always increasing function, and always concave down function.

Since it is a logarithmic function, x>3.

So, x = 3 is vertical asymptote.

I attached the sketch as a picture.

3 0

4 years ago