All categories

3 years ago

Pleeeeeeease help me

Mathematics

1 answer:

kupik [55]3 years ago

6 0

The arrow should be pointing to the negatives

You might be interested in

How can we get System B from System A?<br><br><br> And are they equivalent?

Aliun [14]

What? Where’s the question 11

8 0

3 years ago

PLSSS HELP (2 + 7i) - (6 + 8i) + (10 - 10i)

Montano1993 [528]

(6+9i) is the answer I got.

3 0

3 years ago

Read 2 more answers

5-6x=2+7x step by step plx?

Ivanshal [37]

5-6x=2+7x

+6x +6x

5=2+13x

-2 -2

3=13x

3/3= 13/3

x=4.33

3 0

3 years ago

Ignore the stuff around it ,could someone explain me this?

Ierofanga [76]

Answer:

The lines blue and green are perpendicular

Step-by-step explanation:

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is -1)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

step 1

Find the slope of the blue line

we have the points

(-1,-3) and (0,3)

substitute

$m=\frac{3+3}{0+1}$

$m=\frac{6}{1}=6$

step 2

Find the slope of the red line

we have the points

(3,-3) and (4,2)

substitute

$m=\frac{2+3}{4-3}$

$m=\frac{5}{1}=5$

step 3

Find the slope of the green line

we have the points

(-4,-1) and (2,-2)

substitute

$m=\frac{-2+1}{2+4}$

$m=-\frac{1}{6}$

step 4

Compare the slopes

Blue line ----> $m=6$

Red line ----. $m=5$

Green line ---> $m=-\frac{1}{6}$

The slope of the blue line and the green line are opposite reciprocal ( their product is equal to -1)

therefore

The lines blue and green are perpendicular

7 0

3 years ago

Sin(60-theta)sin(60+theta)

ehidna [41]

Step-by-step explanation:

$\sin(60 - \theta) \sin(60 + \theta) \\ = \{ \sin(60) \cos( \theta) - \sin( \theta) \cos(60) \} \{ \sin(60) \cos( \theta) + \cos(60) \sin( \theta) \} \\ = \{ \frac{ \sqrt{3} }{2} \cos( \theta) - \frac{1}{2} \sin( \theta) \} \{ \frac{ \sqrt{3} }{2} \cos( \theta) + \frac{1}{2} \sin( \theta) \} \\ \\$

from difference of two squares:

${ \boxed{(a - b)(a + b) = ( {a}^{2} - {b}^{2} ) }}$

therefore:

$= \{ {( \frac{ \sqrt{3} }{2}) }^{2} { \cos }^{2} \theta \} - \{ {( \frac{ \sqrt{3} }{2} )}^{2} { \sin }^{2} \theta \} \\ \\ = \frac{3}{4} { \cos }^{2} \theta - \frac{3}{4} { \sin}^{2} \theta$

factorise out ¾ :

$= \frac{3}{4} ( { \cos }^{2} \theta - { \sin}^{2} \theta) \\ \\ = { \boxed{ \frac{3}{4} \cos(2 \theta) }}$

3 0

3 years ago

Read 2 more answers