Area of the rectangle: 112
Step-by-step explanation:
Picture is missing: find it in attachment.
The area of a triangle is given by

where
b is the length of the base
h is the height
For triangle ABE, we know that
A = 40 (area)
h = 8 (height)
So we can find the length of the base AE:

Now we observe that the base of the rectangle, AD, is the sum of AE and DE, therefore:

We also know the height of the rectangle AB is 8, and that the area of a rectangle is

where b is the base and h the height. Therefore, the area of this rectangle is

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<h3>
Answer: 375</h3>
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Work Shown:
a = 300 = first term
r = 60/300 = 0.2 = common ratio
We multiply each term by 0.2, aka 1/5, to get the next term.
Since -1 < r < 1 is true, we can use the infinite geometric sum formula below
S = a/(1-r)
S = 300/(1-0.2)
S = 300/0.8
S = 375
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As a sort of "check", we can add up partial sums like so
- 300+60 = 360
- 300+60+12 = 360+12 = 372
- 300+60+12+2.4 = 372+2.4 = 374.4
- 300+60+12+2.4+0.48 = 374.4+0.48 = 374.88
and so on. The idea is that each time we add on a new term, we should be getting closer and closer to 375. I put "check" in quotation marks because it's probably not the rigorous of checks possible. But it may give a good idea of what's going on.
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Side note: If the common ratio r was either r < -1 or r > 1, then the terms we add on would get larger and larger. This would mean we don't approach a single finite value with the infinite sum.
You would use unit rate for this. If Austin makes $209 in 11 hours, then he makes 209/11 = 19 dollars in 1 hour.
Then, we can make a proportion:
$19/1 hour = $152/ x hours
Cross multiply:
152 = 19x
Solve for x to get:
x = 8 hours.
3.73 rounded to the nearest whole number would be 4
If you have a unit circle, it means that you have a circle of radius r = 1 unit. So if we consider the scalene triangle of the figure, its vertical side is the sine of the theta angle and its horizontal side is the cosine of the theta angle. Then to find the hypotenuse of the triangle you use the Pythagorean theorem, thus relating the sine and cosine measures. I attach the solution.