(Will get the brainest answer if answer is correct.)

( 6 , -10 ) y (-15 , 15 ) Finding slope from two points :)

svet-max [94.6K]

Answer:

-25/21

Step-by-step explanation:

m=(y2-y1)/(x2-x1)=(15-(-10))/(-15-6)=(15+10)/-21=25/-21

5 0

2 years ago

What is the length (magnitude) of the vector (5 -2) a. √29 b. 5√5 c. 5 d.√2

malfutka [58]

Answer:

a. √29

Step-by-step explanation:

The formula for the magnitude of a vector is magnitude = sqrt(x^2 + y^2).

For vector (5, -2):

magnitude = sqrt(5^2 + -2^2)

magnitude = sqrt(25 + 4)

magnitude = sqrt(29)

Therefore, the answer is √29.

Hope this helped :D

8 0

2 years ago

Cos^2x+cos^2(120°+x)+cos^2(120°-x) i need this asap. pls help me

o-na [289]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

= - cos60cosx - sin60sinx

= - $\frac{1}{2}$ cosx - $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos² (120 + x)

= $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

= -cos60cosx + sin60sinx

= - $\frac{1}{2}$ cosx + $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos²(120 - x)

= $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------------

Putting it all together

cos²x + $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x + $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

= cos²x + $\frac{1}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ (cos²x + sin²x) = $\frac{3}{2}$

5 0

2 years ago

Write an equation that is perpendicular to and 3x+y=3 whose y-intercept 5. Anyone please help me, no link just explain how to do

Liula [17]

Answer:

$y = \frac{1}{3}x + 5$

Step-by-step explanation:

By definition, two lines are perpendicular if and only if their slopes are negative reciprocals of each other: $m = - \frac{1}{m_{2} }$ , or equivalently, $m_{1} * m_{2} = -1$ .

Given our linear equation 3x + y = 3 (or y = -3x + 3):

We can find the equation of the line (with a y-intercept of 5) that is perpendicular to y = -3x + 3 by determining the negative reciprocal of its slope, -3, which is $\frac{1}{3}$ .