A quadrilateral has congruent side lengths. 2 opposite angles have measures of 100 degrees. The other 2 opposite angles have mea
2 answers:
Answer:
Option A.
Step-by-step explanation:
It is given that a quadrilateral has congruent side lengths. It means the quadrilateral is either square of rhombus.
Each interior angles of a square is a right angle.
2 opposite angles have measures of 100 degrees. The other 2 opposite angles have measures of 70 degrees and (5 x) degrees.
Since the interior angles of the given quadrilateral are not right angles, therefore it must be a rhombus.
Opposite angles of a rhombus are congruent.
Divide both sides by 5.
The value of x is 14.
Therefore, the correct option is A.
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A) Answer: A. 22.5 cm, 30 cm, and 37.5 cm
Setting the unity measurement as x, the three sides will be 3·x, 4·x, and 5·x. The sum of the three sides will be (3 + 4 + 5)·x = 12·x; this corresponds to the perimeter, therefore:
12·x = 90 cm
We can solve for x:
x = 90 / 12 = 7.5 cm
Hence, the sides will be:
7.5 · 3 = 22.5 cm
7.5 · 4 = 30 cm
7.5 · 5 = 37.5 cm
This situation is represented in option A (which is also the only option whose sides add up to 90).
B) Answer: C. never true
We have:
6 + 8x - 9 = 11x + 14 - 3x
Bring all the terms with x on the left, by subtracting from both sides 11x and -3x, and all the numbers on the right by subtrcting from both sides 6 and -9:
6 + 8x - 9 - 11x - (-3x) - 6 - (-9) = 11x + 14 - 3x - 11x - (-3x) - 6 - (-9)
Cancel out the opposite terms from each side:
8x - 11x + 3x = 14 - 6 + 9
Combine likely terms:
0x = 17
Now you should ask yourself: what number multiplied by zero gives 17 as a result? The answer is none because a number multiplied by zero always equals zero.
Hence, the statement is never true.
C) Answer:
We have:
5r + 4 ≤ 5
Subtract 4 from both sides:
5r + 4 - 4 ≤ 5 - 4
Combine likely terms:
5r ≤ 1
Divide both sides by 5:
To graph the set of solutions, you need to set:
y₁ = 5x + 4 and
y₂ = +5
Plot these two lines:
y₂ is a horizontal line at a height of 5;
y₁ is a line passing through (0, 4) and (1, 9)
The set of solutions will be the part of the graph in which y₁ lies underneath y₂ (see picture attached), that happens for .
