You would add 3 on both sides
3x - 3 + 3 = 13 + 3
3x = 16
Answer:
See step-by-step explanation
Step-by-step explanation:
Question 3: (4, -3)
Solving through elimination: Multiply the first equation by -3.
-3x+6y=-30
3x-5y = 27
Add. The x drops out.
y= -3
Substitute -3 in for y.
x-2y=10
x-2(-3)=10
x+6=10
x=4
Question 4: (2,5)
Solve through elimination: Multiply the second equation with 3.
3/2= 6/4
15/4x=7.5
x=2
3/2 (2) + y = 8
3+ y =8
y=5
Question 5: (6,4)
Use your graph to graph.
Slope of -1/3
Y-intercept of 6 (0,6)
X-intercept of 18 (18,0)
Slope of 3/2
Y-intercept of -5 (0,-5)
X-intercept of (,0)
Question 6: (5, -7)
Solve by substitution:
Words of encouragement:
Good luck!
Two straight lines or edges intersect
Use the chain rule:
<em>y</em> = tan(<em>x</em> ² - 5<em>x</em> + 6)
<em>y'</em> = sec²(<em>x</em> ² - 5<em>x</em> + 6) × (<em>x</em> ² - 5<em>x</em> + 6)'
<em>y'</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)
Perhaps more explicitly: let <em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6, so that
<em>y(x)</em> = tan(<em>x</em> ² - 5<em>x</em> + 6) → <em>y(u(x))</em> = tan(<em>u(x)</em> )
By the chain rule,
<em>y'(x)</em> = <em>y'(u(x))</em> × <em>u'(x)</em>
and we have
<em>y(u)</em> = tan(<em>u</em>) → <em>y'(u)</em> = sec²(<em>u</em>)
<em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6 → <em>u'(x)</em> = 2<em>x</em> - 5
Then
<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>u</em>)
or
<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)
as we found earlier.
60 degrees
Right angles add up to 90 so 90-30=60