To find the area of his exclusion zone you would need to understand that a triangle with dimensions of 3, 4, and 5 represent a right triangle.
This means the exclusion zone would be applied to the base and the height of the triangular space.
You would add 2 km to the 3 km, and 2 km to the 4 km to create a new height of 5 km and a new base of 6 km.
Please see the attached picture to understand this.
You will find the area of the total space created by the new triangle and subtract the space represented by the original triangle to find the area of the exclusion zone.
(1/2 x 6 x 5) - (1/2 x 4 x 3)
15 km² -6 km² equals 9 km².
The exclusion space is 9 km².
Answer:
no, there is not enough information to use AAS congruence theorem to prove the triangles are congruent
Answer:
A, B, C
Step-by-step explanation:
Step 1: "AB ≅ DE, AC ≅ DF, and ∠A ≅ ∠D"
A. Given.
This is the information that was given in the problem statement.
Step 2: "ΔABC ≅ ΔDEF"
B. Side-Angle-Side Postulate (SAS)
The SAS postulate says that if two triangles have a pair of congruent angles between two pairs of congruent sides, then the triangles must be congruent. From the previous step, we can conclude the triangles are congruent.
Step 3: "∠C ≅ ∠F"
C. Corresponding parts of congruent triangles are congruent (CPCTC)
In Step 2, we established the triangles are congruent. So now we can conclude that the corresponding angles are congruent.
Answer:
Step-by-step explanation:
The number of samples is large(greater than or equal to 30). According to the central limit theorem, as the sample size increases, the distribution tends towards normal. The formula is
z = (x - µ)/(σ/√n)
Where
x = sample mean
µ = population mean
σ = population standard deviation
n = number of samples
From the information given,
µ = 22199
σ = 5300
n = 30
the probability that a senior owes a mean of more than $20,200 is expressed as
P(x > 20200)
Where x is a random variable representing the average credit card debt for college seniors.
For n = 30,
z = (20200 - 22199)/(5300/√30) =
- 2.07
Looking at the normal distribution table, the probability corresponding to the z score is 0.0197
P(x > 20200) = 0.0197
Answer:
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