How do you do this? I'll give brainliest, please help.

Mathematics

1 answer:

Dmitriy789 [7]3 years ago

6 0

Answer:

Domain is all values of X or (-∞,∞)

Range is all the possible values of Y. Since it goes up to 3, and then it goes down, it's all values less than or equal to y, or (-∞,3] or y ≤ 3 (depending on how you need to enter the response.

Step-by-step explanation:

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Twice the difference of a number + -3 is 4 find the number

NNADVOKAT [17]

Answer:

The number is 5

Step-by-step explanation:

If we need to find the hidden number, Which for simplicity's sake we will denote as "n", lets first write out the equation.

"2 x [n + (-3)] = 4"

We can work through this problem backwards. if we need two times the sum of "n" and (-3) to get to 4, the only number that can be multiplied by 2 to get a product of 4 is 2, so "n" + (-3) must equal 2. so, From here, we can figure out that since 5 + (-3) equals 2, "n" Must equal 5. so, if we plug in 5 for "n" the equation looks like this:

"2 x [5 + (-3)] = 4"

So, since the equation makes sense, the answer is, "n = 5"

7 0

3 years ago

A triangle ABC has its vertices at A(-2, -3), B(2, 1), and C(5,-1).

maksim [4K]

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill ~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) ~\hfill AB=\sqrt{( 2- (-2))^2 + ( 1- (-3))^2} \\\\\\ AB=\sqrt{(2+2)^2+(1+3)^2}\implies AB=\sqrt{32}\implies \boxed{AB=4\sqrt{2}} \\\\[-0.35em] ~\dotfill$

$B(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad C(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) ~\hfill BC=\sqrt{( 5- 2)^2 + ( -1- 1)^2} \\\\\\ BC=\sqrt{3^2+(-2)^2}\implies BC=\sqrt{9+4}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad A(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-3}) ~\hfill CA=\sqrt{( -2- 5)^2 + ( -3- (-1))^2} \\\\\\ CA=\sqrt{(-7)^2+(-3+1)^2}\implies CA=\sqrt{49+(-2)^2}\implies \boxed{CA=\sqrt{53}}$