These angles, 110º and 7xº, are examples of supplementary angles.
Supplementary angles are angles that, when added together, produce a sum of 180º.
This means 110º + 7xº = 180º.
With this equation, 110º + 7xº = 180º, we can isolate the variable and solve for x to answer this question.
110º + 7xº = 180º
The first step to solve for x is to subtract 110º from both sides of the equation.
110º + 7xº - 110º = 7xº
180º - 110º = 70º
7xº = 70º
Since we want 1xº instead of 7xº, we need to divide both sides by 7.
7xº / 7 = xº
70º / 7 = 10º
xº = 10º
Now we have our solution!
The value of x is 10º.
Let's check this to be sure.
7xº, when added to 110º, must equal 180º since 7xº and 110º are supplementary angles.
7xº + 110º = 180º
We found that x is 10º.
7(10)º + 110º = 180º
7 • 10 = 70
70º + 110º = 180º
180º = 180º
This works! That means that x = 10º is in fact the correct answer.
Answer:
x = 10º
Hope this helps!
Answer:
35,136 ml
Step-by-step explanation:
The tank holds 55,000ml. Mr. Clark poured some gas into 3 container, and at the end there the tank had 19,864. To find how much gas was poured out of the tank it is the same as asking how much gas he poured in the 3 containers, all together. We have to find the difference, subtraction:
55,000 - 19,864 = 35,136ml were poured in the 3 containers or amount of gas poured out of the tank.
It starts at 100% - 15% = 85% after the first year. $15,000*0.85 = $12,750
The second year is further reduced 100% - 16% = 84%
$12,750*0.84 = $10,710
Answer:
x-coordinates of relative extrema = 
x-coordinates of the inflexion points are 0, 1
Step-by-step explanation:
Differentiate with respect to x
Differentiate f'(x) with respect to x
At x = ,
We know that if 
then x = a is a point of minima.
So, 
is a point of minima.
For inflexion points:
Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.
So, x-coordinates of the inflexion points are 0, 1
What you want to do here is take this information and plug it into point-slope form. any time you're given a point and a slope, you generally want to plug it into this equation: y - y1 = m(x - x1).
in this equation, m is your slope and (x1, y1) is a given point. plug in your info--slope of -3 and (-5, 2).
y - 2 = -3(x + 5)
that is the equation of your line. however, if you want to graph it, this doesn't really make much sense to you. convert it to slope-intercept form, y = mx + b, by solving for y.
y - 2 = -3(x + 5) ... distribute -3
y - 2 = -3x - 15 ... add 2
y = -3x - 13 is your equation.
to graph this, and any other y = mx + b equation, you want to start with your y-intercept if it's present. your y intercept here is -13, which means the line you wasn't to graph crosses the y-axis at y = -13, or (0, -13). put a point there.
after you've plotted that point, you use your slope to graph more. remember that your slope is "rise over run"--you rise up/go down however many units, you run left/right however many units. if your slope is -3, you want to go down 3 units, then go to the right 1 unit. remember that whole numbers have a 1 beneath them as a fraction. -3/1 is your rise over 1.
