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Answer:
a) ∆ABC ~ ∆EDC by AA similarity
b) ED/AB = 3/4
c) 15 cm
Step-by-step explanation:
a) Two angles in each triangle are the same, so the AA similarity postulate can be used to declare the ∆ABC ~ ∆EDC. (Each triangle includes a right angle and angle C.)
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b) Corresponding sides are ED/AB, DC/BC, EC/AC. The ratio of corresponding sides is ED/BC = (12 cm)/(16 cm) = 3/4.
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c) Using the ratios identified above, we have ...
DC/BC = 3/4 = x/(20 cm)
x = 3/4(20 cm)
x = 15 cm
Answer:
(4,0)
Step-by-step explanation:
It is a linear function so you find the slope by doing y2-y1/x2-x1 and you find the slope to be 1/2. You can then setup an equation using it and graph it to find the next value.
Answer:
The triangle should be 16 inches wide.
Step-by-step explanation:

What we know:
- One of the legs measures 12 inches.
- The hypotenuse is 20 inches.
We need to solve for b. Solving for a while b is 12 would also be acceptable.

We will use the first equation.


The triangle should be 16 inches wide.
One application of volume is determining the density of an object. Assume the object is made of a pure element (eg: gold). If we know the volume (v) of the object, and we know the mass (m), then we can use the formula D = m/v to figure out the density D. Knowing the volume is also handy to determine if the object can fit into a larger space or not. Another application is figuring out how much water is needed to fill up the inner space of the 3D solid (assuming it's hollow on the inside).
The surface area is handy to figure out how much material is needed to cover the outer surface. This material can be paint, paper, metal sheets, or whatever you can think of really. A good example is wrapping a present and the assumption is that there is no overlap.