Answer:
The answers are A, C, and D, hope this helps!
Step-by-step explanation:
I took the assignment and got it right
Divide both sides by -7, and flip the sign: 3 > c
When choosing an interval to use for a frequency table, the low value and the high value of the dat is considered so that the interval refrects a true data size for all the intervals.
Given that a<span>
data set has values ranging from a low of 10 to a high of 52. Using the class limits of 10-19, 20-29, 30-39, 40-49 for a
frequency table will make the last interval not to refrect the true size of the other intervals. i.e. the last interval will be 50 - 52 which has a size of 3, different from the size of 10 the other intervals have.</span>
The far of reaching the wall
24tan71
=69.70106106ft
Answer:
C
Step-by-step explanation:
We have the equation:
Add 10 to both sides to isolate the equation.
This is not factorable*, so we can use the quadratic formula:
In this case, <em>a</em> = 4, <em>b</em> = 5, and <em>c</em> = 10.
Substitute:
Simplify:
Since we cannot take the root of a negative, we have no real solutions.
Our answer is C.
*To factor something in the form of:
We want two numbers <em>p</em> and <em>q</em> such that <em>pq</em> = <em>ac</em> and <em>p</em> + <em>q</em> = <em>b</em>.
Since <em>ac</em> = 4(10) = 40. We need to find two whole numbers that multiply to 40 and add to 5.
No such numbers exist, so the equation is not factorable.