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3 years ago

Which inequality represents all possible solutions of -3(x+5)>12?

Mathematics

1 answer:

IgorC [24]3 years ago

5 0

its A

Step-by-step explanation:

dividing or multiplying a negative number in this will completely change the sign to opp side

-3x + 15 < 12

-3x < -3

x > 1

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trapecia [35]

Answer:

domain: {x ∈ ℝ : x ≤ 5}
range: {y ∈ ℝ : y ≤ -1}

Step-by-step explanation:

Domain

The domain of a function is the set of x values for which the function is defined. Here, the domain is limited by the values of x that make the square root defined. That is, the expression under the radical cannot be negative:

-3x +15 ≥ 0

15 ≥ 3x . . . . . . add 3x

5 ≥ x . . . . . . . . divide by 3

x ≤ 5 . . . . . . . . put x on the left (swap sides)

The rest of the notation in the domain expression simply says x is a real number.

domain: {x ∈ ℝ : x ≤ 5} . . . . . . matches the first choice

Range

The range of a function is the set of values that f(x) can have. We know the square root can be zero or any positive number. When it is zero, f(x) = -1.

When it is a positive number, that value is multiplied by -4 and added to -1, so f(x) is a number more negative than -1. Then the range of the function is all numbers -1 and below:

range: {y ∈ ℝ : y ≤ -1} . . . . . . matches the last choice

_____

Comment on domain/range problems

When working domain and range problems, it works well to have a good understanding of the domain and range limitations of the functions we usually work with: polynomials, square root, logarithm, trig functions, exponential functions. Domain and range problems generally involve combinations of these or ratios of combinations of these.

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3 years ago

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Inga [223]

Answer:

$y=-\sqrt{3}x+2$

Step-by-step explanation:

We want to find the equation of a straight line that cuts off an intercept of 2 from the y-axis, and whose perpendicular distance from the origin is 1.

We will let Point M be (x, y). As we know, Point R will be (0, 2) and Point O (the origin) will be (0, 0).

First, we can use the distance formula to determine values for M. The distance formula is given by:

$\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Since we know that the distance between O and M is 1, d=1.

And we will let M(x, y) be (x₂, y₂) and O(0, 0) be (x₁, y₁). So:

$\displaystyle 1=\sqrt{(x-0)^2+(y-0)^2}$

Simplify:

$1=\sqrt{x^2+y^2}$

We can solve for y. Square both sides:

$1=x^2+y^2$

Rearranging gives:

$y^2=1-x^2$

Take the square root of both sides. Since M is in the first quadrant, we only need to worry about the positive case. Therefore:

$y=\sqrt{1-x^2}$

So, Point M is now given by (we substitute the above equation for y):

$M(x,\sqrt{1-x^2})$

We know that Segment OM is perpendicular to Line RM.

Therefore, their slopes will be negative reciprocals of each other.

So, let’s find the slope of each segment/line. We will use the slope formula given by:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Segment OM:

For OM, we have two points: O(0, 0) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{OM}=\frac{\sqrt{1-x^2}-0}{x-0}=\frac{\sqrt{1-x^2}}{x}$

Line RM:

For RM, we have the two points R(0, 2) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{RM}=\frac{\sqrt{1-x^2}-2}{x-0}=\frac{\sqrt{1-x^2}-2}{x}$

Since their slopes are negative reciprocals of each other, this means that:

$m_{OM}=-(m_{RM})^{-1}$

Substitute:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=-\Big(\frac{\sqrt{1-x^2}-2}{x}\Big)^{-1}$

Now, we can solve for x. Simplify:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=\frac{x}{2-\sqrt{1-x^2}}$

Cross-multiply:

$x(x)=\sqrt{1-x^2}(2-\sqrt{1-x^2})$

Distribute:

$x^2=2\sqrt{1-x^2}-(\sqrt{1-x^2})^2$

Simplify:

$x^2=2\sqrt{1-x^2}-(1-x^2)$

Distribute:

$x^2=2\sqrt{1-x^2}-1+x^2$

So:

$0=2\sqrt{1-x^2}-1$

Adding 1 and then dividing by 2 yields:

$\displaystyle \frac{1}{2}=\sqrt{1-x^2}$

Then:

$\displaystyle \frac{1}{4}=1-x^2$

Therefore, the value of x is:

$\displaystyle \begin{aligned}\frac{1}{4}-1&=-x^2\\-\frac{3}{4}&=-x^2\\ \frac{3}{4}&=x^2\\ \frac{\sqrt{3}}{2}&=x\end{aligned}$

Then, Point M will be:

$\begin{aligned} \displaystyle M(x,\sqrt{1-x^2})&=M(\frac{\sqrt{3}}{2}, \sqrt{1-\Big(\frac{\sqrt{3}}{2}\Big)^2)}\\M&=(\frac{\sqrt3}{2},\frac{1}{2})\end{aligned}$

Therefore, the slope of Line RM will be:

$\displaystyle \begin{aligned}m_{RM}&=\frac{\frac{1}{2}-2}{\frac{\sqrt{3}}{2}-0} \\ &=\frac{\frac{-3}{2}}{\frac{\sqrt{3}}{2}}\\&=-\frac{3}{\sqrt3}\\&=-\sqrt3\end{aligned}$

And since we know that R is (0, 2), R is the y-intercept of RM. Then, using the slope-intercept form:

$y=mx+b$

We can see that the equation of Line RM is:

$y=-\sqrt{3}x+2$

6 0

3 years ago

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PLEASE HELP , will reward brainliest

Bess [88]

The answer is 0 because If two events are mutually exclusive, then the probability of either occurring is the sum of the probabilities of each occurring.

Hope that Helps :)

5 0

3 years ago

Read 2 more answers