Part A.
You need two equations with the same slope and different y-intercepts.
Their graph is parallel lines. Since the lines do not intersect, there is no solution.
y = 2x + 2
y = 2x - 2
Part B.
We use the first equation as above. For the second equation, we use an equation with different slope. Two lines with different slopes always intersect.
y = 2x + 2
y = -2x - 2
In the second equation, y = -2x - 2. We now substitute -2x - 2 for y in the first equation.
-2x - 2 = 2x + 2
-4x = 4
x = -1
Now substitute -1 for x in the first equation to find y.
y = 2x + 2
y = 2(-1) + 2
y = -2 + 2
y = 0
Solution: x = -1 and y = 0
Answer:
6 toppings
Step-by-step explanation:
The t in this equation represents the amount of toppings.
22 + 0.65t = 26
-22 -22
0.65t = 4
0.65t/0.65 = 4/0.65
t = 6
They could get 6 toppings and the total would be $25.90.
The point-slope formula is y-y1=m(x-x1). You will then substitute into this equation. The answer will be y-2=-8(x+3).
We know that
<span>Since the focus and vertex are above and below each other, rather than side by side, I know that this ellipse must be taller than it is wide.
</span>Then
a²<span> will go with the </span>y<span> part of the equation
</span>Also, since the focus is 8 <span>units below the center, then </span><span>c = 8
</span>since the vertex is 17<span> units above, then </span><span>a = 17
</span>The equation b²<span> = a</span>²<span> – c</span>²<span> gives me
</span>b²=17²-8²-----> b²=225
the equation is
y²/a²+x²/b²=1------> y²/289+x²/225=1
the answer isy²/289+x²/225=1see the attached figure
Answer:
Median.
Step-by-step explanation:
We have been given that in 2004, the mean net worth of families in a certain region was $470.2 thousand and the median net worth was $92.3 thousand.
We know that mean and median of a symmetric data set is equal.
We also know that when mean of a data set is greater than median, then the data set has a very large valued outlier.
Since mean net wroth of families is approximately 5 times more than median net wroth of families, this means that some of the families has very high net worth as outliers.
Since the net worth of families has very large outliers, therefore, I would prefer median as the appropriate measure of center as median is not affected by outliers.