Short Answer: The first three are absolutely true. Congruency is maintained if all you are doing is shifting the figure 8 spaces away. Since the figures are congruent, their corresponding parts are equal and the statements made are always true. That's the guiding principle.
A is true. See the short answer.
B is True. You have gone 8 units away. The distance between the old G and the new one is 8 units. This is not a congruency property but it is true because it is given. You haven't altered the shape or rotatated it.
C is true by conguency. You have just shifted things 8 units. You have maintained equality and you have maintained distance between corresponding parts.
That leaves D. If the figures are regular, that will be true. If they are not regular, non corresponding external angles cannot be guaranteed to be true.
No, the expressions –0.5(3x 5) and –1.5x 2.5 are not equivalent.
Using the distributive property to expand the first expression, you would get -1.5x - 2.5. Since the constant term does not have the same sign as the second expression, the two expressions are not equivalent.
Also, if you substitute in a value for x and evaluate each expression, the values will not be equal.
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Let x = amount of 70% juice we use.
Then 60-x = amount of 85% juice we'll use.
We'll set up an equation that is based on how much pure juice comes from each component and how much we need in the final mixture.
So 70% of x plus 85% of (60-x) must equal 80% of 60, or...
0.7x + 0.85(60-x) = 0.8(60)
0.7x + 51 - 0.85x = 48
-0.15x + 51 = 48
-0.15x = -3
x = 20
Then 60 - x = 60 - 20 = 40
Check: 70% of 20 pints = 14 pints of pure juice.
85% of 40 pints = 34 pints of pure juice
That's 48 pints of pure juice in the mix, which matche 80% 0f 60 pints.
You'd need 20 pints of the 70% juice and 40 pints of the 85% juice.
Answer:
Solutions are (3,1) and (4,2)
Step-by-step explanation:
Graph is shown in the attached sheet
Given are two systems of equations and we have to solve them using graph
For graphing let us first prepare table for x and y.
1) 
I line II line
x 0 4.5 3 x 0 2.5 3
y 3 0 1 y -5 0 1
The two lines intersect at (3,1)
Hence solution is (3,1)
--------------------------------------------
2) 
I line II line
x 0 2 4 x 0 6 4
y 0 1 2 y 3 0 2
The two lines intersect at (4,2)
Hence solution is (4,2)
