Answer:
1. -7.5
2. $1
3. 40
Step-by-step explanation:
For number 1, it can be solved by using the PEMDAS method, or see explanation below:
6x - 4x - 36 = 6 - 2x
2x - 36= -2x + 6
4x + 36 = 6
4x = -30
x = -15/2 or -7.5
For number 2, substitute 3 into both equations:
f(x) =1.50(3) + 2.00
and
f(x) = 2.00(3) + 1.50
This would get $6.50 and $7.50, which, if subtracted, gets $1.
For number 3, do something similar to the previous problem. Substitute 3 for x. It would be 5(2^3), or 40.
Hope this helps!
Answer:
68
Step-by-step explanation:
6.8 × 
= 68
Let's go through the choices one by one
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Choice A
If all sides are congruent, then this figure is a rhombus (by definition). If all angles are congruent, then we have a rectangle. Combine the properties of a rhombus with the properties of a rectangle and we have a square.
In terms of "algebra", you can think
rhombus+rectangle = square
Or you can draw out a venn diagram. One circle represents the set of all rhombuses; another circle represents the set of all rectangles. The overlapping region is the set of all squares. The overlapping region is inside both circles at the same time.
So we can rule out choice A. This guarantees we have a square when we want something that isn't a guarantee.
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Choice B
If we had a parallelogram with perpendicular diagonals, then we can prove that we have a rhombus (all four sides congruent). However, we don't know anything about the four angles of this parallelogram. Are they congruent? We don't know. So we can't prove this figure is a rectangle. The best we can say is that it's a rhombus. It may or may not be a rectangle. There isn't enough info about the rectangle & square part.
This is why choice B is the answer. We have some info, but not enough to be guaranteed everytime.
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Choice C
This is a repeat of choice A. Having "all right angles" is the same as saying "all angles congruent". This is because "right angle" is the same as saying "90 degrees". So we can rule out choice C for identical reasons as we did with choice A.
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Choice D
As mentioned before in choice A, if we know that a quadrilateral is a rectangle and a rhombus at the same time, then the figure is also a square. This is always true, so we are guaranteed to have a square. We can cross choice D off the list.
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Once again, the final answer is choice B
Given:
Quadrilateral QRST has diagonals that do NOT bisect each other, are congruent and are not perpendicular.
By checking the options:
Rhombus, rectangle and the square each of them the diagonals bisect each other
So, the answer will be option 4) Isosceles Tra
