All categories

2 years ago

Find the coordinates of a point that is in Quadrant IV and is √45 units away from the point (-4, 1).

Mathematics

1 answer:

exis [7]2 years ago

6 0

Answer:

15-3

Step-by-step explanation:

15x3=45 or it could be 3-15

You might be interested in

And evaluate x(- y + z) for x=-1 y=-2 and z=5

IRISSAK [1]

Multiply x to (-y) then x to (z)
mean x(-y)+ x(z)
plug in the variable given
(-1)(-(-2)+5)
-2-5=-7

5 0

3 years ago

Read 2 more answers

you are given eight golf balls. Seven of the golf balls have the exact same weight, but one ball is slightly lighter(you cannot

Inessa [10]

7 weight exactly the same, therefore the 8th golf ball doesnt weigh the same as the 7. so the 8th ball is slightly lighter

6 0

3 years ago

F(x)=-x^2-4x-6<br> Axis of symmetry and vertex

NikAS [45]

The vertex of the equation is ( -2 , -2 )

The axis of symmetry is x= - 2

5 0

3 years ago

Simplify using trigonometric identities<br> 2sinθ - sin2θ cosθ

Anit [1.1K]

Double angle identity for sine:

$\sin(2x) = 2 \sin(x) \cos(x)$

$\implies 2 \sin(\theta) - \sin(2\theta) \cos(\theta) = 2 \sin(\theta) - 2 \sin(\theta) \cos^2(\theta)$

Factorize the left side.

$2 \sin(\theta) - 2 \sin(\theta) \cos^2(\theta) = 2 \sin(\theta) \left(1 - \cos^2(\theta)\right)$

Pythagorean identity:

$\cos^2(x) + \sin^2(x) = 1$

$\implies 2 \sin(\theta) \left(1 - \cos^2(\theta)\right) = 2 \sin^3(\theta)$

so that

$\boxed{2 \sin(\theta) - \sin(2\theta) \cos(\theta) = 2 \sin^3(\theta)}$

8 0

1 year ago

The vertices of triangle RST are R (-7, 5), S(17, 5), T(5, 0). What is the perimeter of triangle RST?

MaRussiya [10]

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ R(\stackrel{x_1}{-7}~,~\stackrel{y_1}{5})\qquad S(\stackrel{x_2}{17}~,~\stackrel{y_2}{5}) ~\hfill RS=\sqrt{(~~ 17- (-7)~~)^2 + (~~ 5- 5~~)^2} \\\\\\ ~\hfill RS=\sqrt{( 24)^2 + ( 0)^2}\implies \boxed{RS=24} \\\\\\ S(\stackrel{x_1}{17}~,~\stackrel{y_1}{5})\qquad T(\stackrel{x_2}{5}~,~\stackrel{y_2}{0}) ~\hfill ST=\sqrt{(~~ 5- 17~~)^2 + (~~ 0- 5~~)^2}$

$~\hfill ST=\sqrt{( -12)^2 + ( -5)^2}\implies \boxed{ST=13} \\\\\\ T(\stackrel{x_1}{5}~,~\stackrel{y_1}{0})\qquad R(\stackrel{x_2}{-7}~,~\stackrel{y_2}{5}) ~\hfill TR=\sqrt{(~~ -7- 5~~)^2 + (~~ 5- 0~~)^2} \\\\\\ ~\hfill TR=\sqrt{( -12)^2 + (5)^2}\implies \boxed{TR=13} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{\LARGE perimeter}}{24~~ + ~~13~~ + ~~13\implies \text{\LARGE 50}}~\hfill$

3 0

1 year ago