Answer:

Explanation:
[ Divide Both Sides By 12y ]
12xy / 12y = 5z / 12y
- PNW
Answer:

Step-by-step explanation:
Given
--- [Missing in question]
Required
Select equivalent expression

Associative property states that:

In this case:



So, we have:


B. is correct
Step-by-step explanation:
A. Base = <em><u>(-3)</u></em>
B. Exponent = <em><u>5</u></em>
C. (-3)^5
= <em><u>(-3) × (-3) × (-3) × (-3) × (-3) (In expanded notation form)</u></em>
D. Evaluation of equation will be = <em><u>-243</u></em>
Statement 3 and 4 are true as Figures 1 and 2 are not congruent and Figures 1 and 3 are not congruent
<h3>What are Congruent Figures ?</h3>
The figures that are similar in shape and size or can be mapped into one another , such figures are called Congruent Figures.
The graph has been plotted on the basis of given data.
The plot can be seen in the graph attached with the answer.
The statements that are true according to the given data is
Statement 3 and 4 are true as
Figures 1 and 2 are not congruent because figure 1 cannot be mapped onto figure 2 using a sequence of rigid transformations.
Figures 1 and 3 are not congruent because figure 1 cannot be mapped onto figure 3 using a sequence of rigid transformations.
To know more about Congruent Figures
brainly.com/question/12132062
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The height of the mountain estimated by two students is 2,467.98 meters.
<h3>
Height of the mountain </h3>
The height of the mountain is calculated as follows;
tanα = h/( a + b + x)
where;
- x is the distance between end of b and h
tan37 = h/(1850 + 740 + x)
tan37 = h/(2590 + x)
h = tan37(2590 + x)
h =1,951.82 + 0.7536x ---- (1)
tanβ = h/(b + x)
tan60 = h/(740 + x)
h = tan60(740 + x)
h = 1,281.68 + 1.732x ---- (2)
Solve (1) and (2) together
1,951.82 + 0.7536x = 1,281.68 + 1.732x
1,951.82 - 1,281.68 = 1.732x - 0.7536x
670.14 = 0.9784x
x = 684.93 m
h = 1,281.68 + 1.732x
h = 1,281.68 + 1.732(684.93)
h = 2,467.98 m
Thus, the height of the mountain estimated by two students is 2,467.98 meters.
Learn more about angle of elevation here: brainly.com/question/88158
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