All categories

3 years ago

Name two points that are reflections of each other across the y-axis.

Mathematics

1 answer:

zheka24 [161]3 years ago

5 0

Answer:

(1,1), (-1,1)

Step-by-step explanation:

To reflect a point over the x-axis, simply make its x-value negative. So, we can take the sample point of (1,1), make its x-value negative, and get (-1,1).

You might be interested in

Carly spend 10 hours a day working at her job what percentage of her day does she spend at work ?

AlladinOne [14]

10/24 x 100 = 41.66
Approximately 42% of her day

3 0

3 years ago

Read 2 more answers

Point A is located at (−2, −6), and D is located at (−6, 8). Find the coordinates of the point that lies halfway between A and D

Afina-wow [57]

Answer:

(-4,1)

Step-by-step explanation:

Point A is located at (−2, −6), and D is located at (−6, 8).

We need to find the midpoint of A and D

Mid point formula is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Point A is (-2,-6) that is (x1,y1)

Point D is (-6,8) that is (x2, y2)

plug in the values in the formula

$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

$(\frac{-2-6}{2}, \frac{-6+8}{2})$

$(\frac{-8}{2}, \frac{2}{2})$

Mid point is (-4, 1)

5 0

3 years ago

Read 2 more answers

Can you please help me with this question

patriot [66]

7 guests
It costs 5 dollars per guest, so divide 35 by 5

3 0

2 years ago

Read 2 more answers

Find all values of t such that t+4/t+5 = t-5/2t

ivolga24 [154]

Answer:

t cannot be -5 or 0 because these cause a 0 in the denominator and you cant divide by 0 that I'm aware of

3 0

3 years ago

Which of the following would be an acceptable first step in simplifying the expression sinx/1-sinx

Paha777 [63]

$\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
\textit{also recall that }sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-------------------------------$

$\bf \cfrac{sin(x)}{1-sin(x)}\implies \cfrac{sin(x)}{1-sin(x)}\cdot \cfrac{1+sin(x)}{1+sin(x)}\implies \stackrel{first~step}{\cfrac{sin(x)[1+sin(x)]}{[1-sin(x)][1+sin(x)]}}
\\\\\\
\cfrac{sin(x)[1+sin(x)]}{1^2-sin^2(x)}\implies \cfrac{sin(x)[1+sin(x)]}{cos^2(x)}
\\\\\\
\cfrac{sin(x)+sin^2(x)}{cos^2(x)}\implies \cfrac{sin(x)}{cos^2(x)}+ \cfrac{sin^2(x)}{cos^2(x)}
\\\\\\
\cfrac{sin(x)}{cos(x)}\cdot \cfrac{1}{cos(x)}+\cfrac{sin^2(x)}{cos^2(x)}\implies tan(x)sec(x)+tan^2(x)
\\\\\\
tan(x)[sec(x)+tan(x)]$

8 0

3 years ago

Read 2 more answers