Answer:
x = lillies
y = tulips
x + y = 13
1.25x + .90y = 14.85
-1.25x - 1.25y = -16.25
1.25x + .90y = 14.85
-0.35y = -1.40
y = 4 tulips
x + 4 = 13
x = 9 lillies
(9 lillies, 4 tulips)
Step-by-step explanation:
Answer:
≥ 999,999,999
Step-by-step explanation:
that is the less than or equal to sign
hope this helps
Answer:
(-4, -1)
Step-by-step explanation:
-3x - 8y = 20; y = 5x + 19
y = 5x + 19; -3x - 8y = 20
y = 5x + 19
-3x - 8y = 20
-3x - 8(5x + 19) = 20
-43x - 152 = 20
-43x - 152 + 152 = 20 + 152
-43x = 172
-43x / -43 = 172 / -43
x = -4
y = 5x + 19
y = (5)(-4) + 19
y = -1
<em>4℉.</em>
What we know about Degrees is that there is a<em> </em><u><em>Positive</em></u> type and a <u><em>negative</em></u> type.
(i.e: 30℉ is <u><em>positive</em></u> and -30℉ is <u><em>negative</em></u>.)
If the temperature was -4℉ at 7AM, then it is negative. If it goes up by an amount that is more than 4 then that negative will go up to a positive temperature. In this case: At 9AM it was 8° <u><em>warmer</em></u>.
<u><em>Warmer</em></u><em> is a </em><u><em>keyword</em></u><u>.</u> If it is warmer by an amount, Negative temperatures <u><em>will go up to a positive</em></u> and positive temperature <u><em>will just go up</em></u>. If it gets cooler, negative temperatures <u><em>will go down further</em></u> and positive temperatures <u><em>will go down to a negative</em></u>.
So lets work out this problem with our newfound knowledge.
-4° F at 7AM
8° warmer at 9AM
-4 + 8 = 4.
<em>The temperature was 4° at 9AM.</em>
-Snooky
Answer:
Step-by-step explanation:
The equation 
represents the discriminant of a quadratic. It is the part taken from under the radical in the quadratic formula.
For any quadratic:
- If the discriminant is positive, or greater than 0, the quadratic has two solutions
- If the discriminant is equal to 0, the quadratic has one distinct real solution (the solution is repeated).
- If the discriminant is negative, or less than 0, the quadratic has zero solutions
In the graph, we see that the equation intersects the x-axis at two distinct points. Therefore, the quadratic has two solutions and the discriminant must be positive. Thus, we have 
.
