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2 years ago

What is the domain of the function y = l n (StartFraction negative x 3 Over 2 EndFraction).

Mathematics

1 answer:

Ymorist [56]2 years ago

3 0

The domain of the provide function where the function is defined is x less than the number three.

<h3>What is the domain of the function?</h3>

Domain of a function is the set of all the possible input values which are valid for that function.

The function given in the problem is,

$y=\ln\left(\dfrac{-x+3}{1}\right)$

The above function is the function of logarithm. For the logarithmic function, it is defined for all the real numbers except 0.

Thus, the function should be greater than zero as,

$\left(\dfrac{-x+3}{1}\right) > 0\\(-x+3) > 0\\-x > -3$

Change the sign of both side, by changing the inequality,

$x < 3$

Thus, the domain of the provide function where the function is defined is x less than the number three.

Learn more about the domain of the function here;

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5.4p+13.1=−2.6p+3.5 <br> p=?

mote1985 [20]

Answer:

P= - 16/11 or P= - 1 5/11 or P= - 1.45455

Step-by-step explanation:

Just trust my answer

7 0

3 years ago

What decimal can be added to 0.0(54) to get 0.1(54)? (the parenthesis mean repeating)

son4ous [18]

Add 0.1 to 0.0(54) to get 0.1(54).

5 0

3 years ago

Read 2 more answers

Please help me now ! Thank you

jarptica [38.1K]

First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now

2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4

At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real.

Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result.

For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196

No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.

So x^2 = -4 has no real solutions. Taking the square root of both sides leads to

x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
which are complex non-real values

5 0

3 years ago

Suppose that salaries for recent graduates of one university have a mean of $25,700 with a standard deviation of $1350. Using Ch

sergeinik [125]

Answer:

75% of the data will reside in the range 23000 to 28400.

Step-by-step explanation :

To find the range of values :

We need to find the values that deviate from the mean. Since we want at least 75% of the data to reside between the range therefore we have,

$1 - \frac{1}{k^2} =\frac{75}{100}$

Solving this, we would get k = 2 which shows the value one needs to find lies outside the range.

Range is given by : mean +/- (z score) × (value of a standard deviation)

⇒ Range : 25700 +/- 2 × 1350

⇒ Range : (25700 - 2700) to (25700 + 2700)

Hence, 75% of the data will reside in the range 23000 to 28400.

3 0

3 years ago

HELP! Find the value of sin 0 if tan 0 = 4; 180 < 0< 270

BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

$\large \boxed{sin \theta = tan \theta cos \theta} \\ \large \boxed{tan^{2} \theta + 1 = {sec}^{2} \theta}$

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

$\large{ {4}^{2} + 1 = {sec}^{2} \theta } \\ \large{16 + 1 = {sec}^{2} \theta } \\ \large{ {sec}^{2} \theta = 17}$

As we know, sec²θ = 1/cos²θ.

$\large \boxed{sec \theta = \frac{1}{cos \theta} } \\ \large \boxed{ {sec}^{2} \theta = \frac{1}{ {cos}^{2} \theta} }$

And thus,

$\large{ {cos}^{2} \theta = \frac{1}{17}} \\ \large{cos \theta = \frac{ \sqrt{1} }{ \sqrt{17} } } \\ \large{cos \theta = \frac{1}{ \sqrt{17} } \longrightarrow \frac{ \sqrt{17} }{17} }$

Since the given domain is 180° < θ < 360°. Thus, the cosθ < 0.

$\large{cos \theta = \cancel\frac{ \sqrt{17} }{17} \longrightarrow cos \theta = - \frac{ \sqrt{17} }{17}}$

Then use the Identity of sinθ = tanθcosθ to find the sinθ.

$\large{sin \theta = 4 \times ( - \frac{ \sqrt{17} }{17}) } \\ \large{sin \theta = - \frac{4 \sqrt{17} }{17} }$

Answer

sinθ = -4sqrt(17)/17 or A choice.

4 0

3 years ago