The third one. The one that goes long box, short, short, long. That's because the first box isn't covered, and they next two are about half boxes showing and the last one is a full box again.
Answer:
y + b > 15
Step-by-step explanation:
given data
positive integers = x, y, a, b
x divided by y then remainder = 6
a divided by b then remainder = 9
to find out
possible value for y + b
solution
we know that here when x divided by y then remainder is 6
its mean y is greater than 6 and when a divided by b then remainder is 9
so its mean b is greater than 9
and we know remainder is less than divisor
so here y + b must be greater than = 6 + 9
y + b > 15
Its 8/10 or 12/15 because 4 times 2 is 8 and 5 times 2 is ten
Option D:
ΔCAN ≅ ΔWNA by SAS congruence rule.
Solution:
Given data:
m∠CNA = m∠WAN and CN = WA
To prove that ΔCAN ≅ ΔWNA:
In ΔCAN and ΔWNA,
CN = WA (given side)
∠CNA = ∠WAN (given angle)
NA = NA (reflexive side)
Therefore, ΔCAN ≅ ΔWNA by SAS congruence rule.
Hence option D is the correct answer.
The four inequalities that can be used to find the solution of 3 ≤ |x + 2| ≤ 6 is x + 2 ≤ 6, x + 2 ≥ -6, x + 2 ≥ 3 and x + 2 ≤ -3
<h3>What is an
equation?</h3>
An equation is an expression that shows the relationship between two or more variables and numbers.
Given the inequality:
3 ≤ |x + 2| ≤ 6
Hence:
x + 2 ≤ 6, -(x + 2) ≤ 6, 3 ≤ x + 2 and 3 ≤ -(x + 2)
This gives:
x + 2 ≤ 6, x + 2 ≥ -6, x + 2 ≥ 3 and x + 2 ≤ -3
The four inequalities that can be used to find the solution of 3 ≤ |x + 2| ≤ 6 is x + 2 ≤ 6, x + 2 ≥ -6, x + 2 ≥ 3 and x + 2 ≤ -3
Find out more on equation at: brainly.com/question/2972832
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