Find one value of xxx that is a solution to the equation: (2x+3)^2-4(2x+3)-12=0(2x+3) 2 −4(2x+3)−12=0
1 answer:
Answer:
x = -5/2 or 3/2
Step-by-step explanation:
If we let z = 2x+3, the equation can be written ...
... z² -4z -12 = 0
and can be factored as
... (z -6)(z +2) = 0
Then the solutions are z = -2, z = 6.
This means the values of x that are solutions are ...
... 2x +3 = {-2, 6} . . . . work with both possibilities at once
... 2x = {-5, 3} . . . . . . . subtract 3
... x = {-5/2, 3/2} . . . . .divide by 2
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Answer:
x = 12
∠A = 144
Step-by-step explanation:
+ We have: ∠A = ∠C (alternate interior)
∠D = ∠C ( vertically opposite angles)
=> ∠A = ∠C (because = ∠C)
=> 10x + 24 = 6x + 72
=> 10x - 6x = 72 - 24
4x = 48
x = 12
+ ∠A = 10x + 24
∠A = 10 x 12 + 24
∠A = 144
Answer:
Minimum = 14
Median = 32
Maximum = 65
Step-by-step explanation:
<u>Step 1: </u>Put your numbers in ascending order (from smallest to largest). For this particular data set, the order is
14, 18, 28, 29, 32, 44, 48, 50, 65
<u>Step 2:</u> Find the minimum and maximum for your data set. Now that your numbers are in order, this should be easy to spot.
Minimum = 14
Maximum = 65
<u>Step 3: </u>Find the median. The median is the middle number. The middle number in your list is 32, so
Median = 32
<u>Step 4: </u>Place parentheses around the numbers above and below the median.
(14, 18, 28, 29), 32, (44, 48, 50, 65)
<u>Step 5: </u>Find Q1 and Q3. Q1 can be thought of as a median in the lower half of the data, and Q3 can be thought of as a median for the upper half of data.
First quartile
Third quartile
<u>Step 6:</u> Write down your summary found in the above steps.
Minimum = 14
Median = 32
Maximum = 65