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3 years ago

Please help on two math questions. I don’t understand how to do them.

Mathematics

1 answer:

Sidana [21]3 years ago

3 0

When you represent intervals on the number line, you're including full dots, excluding empty dots, and you're considering numbers highlighted by the line.

In the first case, you've highlighted everything before -2 (full dot, thus included), and everything after 1 (empty dot, excluded). So, the set would be

$x\leq -2 \lor\ x>1$

or, in interval notation,

$(-\infty,-2]\cup (1,\infty)$

In the second case, you are looking for all numbers between -3 and 5. This interval is symmetric with respect to 1: you're considering all numbers that are at most 4 units away from 1, both to the left and to the right.

This means that the difference between your numbers at 1 must be at most 4, which is modelled by

$|x-1|\leq 4$

where the absolute values guarantees that you'll pick numbers to the left and to the right of 1.

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Find the locus of a point such that the sum of its distance from the point ( 0 , 2 ) and ( 0 , -2 ) is 6.

jok3333 [9.3K]

Answer:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Step-by-step explanation:

We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.

First, we will need the distance formula, given by:

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

Let the point on the locus be P(x, y).

So, the distance from P to (0, 2) will be:

$\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}$

And, the distance from P to (0, -2) will be:

$\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}$

So sum of the two distances must be 6. Therefore:

$d_1+d_2=6$

Now, by substitution:

$(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6$

Simplify. We can subtract the second term from the left:

$\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}$

Square both sides:

$(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)$

We can cancel the x² terms and continue squaring:

$y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4$

We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:

$-8y=36-12\sqrt{x^2+(y+2)^2}$

We can divide both sides by -4:

$2y=-9+3\sqrt{x^2+(y+2)^2}$

Adding 9 to both sides yields:

$2y+9=3\sqrt{x^2+(y+2)^2}$

And, we will square both sides one final time.

$4y^2+36y+81=9(x^2+(y^2+4y+4))$

Distribute:

$4y^2+36y+81=9x^2+9y^2+36y+36$

The 36y will cancel. So:

$4y^2+81=9x^2+9y^2+36$

Subtracting 4y² and 36 from both sides yields:

$9x^2+5y^2=45$

And dividing both sides by 45 produces:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.

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Answer:

58/100

50 X 2 = 100

29 X 2 = 58

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Which ordered pairs are solutions to the system? Check all of the boxes that apply.

77julia77 [94]

Answer:

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Step-by-step explanation:

I just did this question

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Trigonometric equations<br><br> 5sinx = 3sinx + square root of 2

Lera25 [3.4K]

$\qquad\qquad\huge\underline{{\sf Answer}}♪$

Let's solve for x ~

$\qquad \sf \dashrightarrow \:5 \sin(x) = 3 \sin(x) + \sqrt{2}$

$\qquad \sf \dashrightarrow \:5 \sin(x) - 3 \sin(x) = \sqrt{2}$

$\qquad \sf \dashrightarrow \:2 \sin(x) = \sqrt{2}$

$\qquad \sf \dashrightarrow \: \sin(x) = \sqrt{2} \div 2$

$\qquad \sf \dashrightarrow \: \sin(x) = \frac{1}{ \sqrt{2} }$

$\qquad \sf \dashrightarrow \:x = 45 \degree \: \: or \: \: \frac{\pi}{4} \: rad$

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Line m passes through C(-2, 0) and D(1, -3). What is the equation of the line m in standard form?

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