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Serga [27]
3 years ago
12

Answer this question to get marked as barinliest!!!!!

Mathematics
1 answer:
GrogVix [38]3 years ago
8 0

Answer:

D- 30 square units

Step-by-step explanation:

5 times 12 =60

60/2= 30

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100-point Question!!!
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Answer:

Two or more independent functions (say f(x) and g(x)) can be combined to generate a new function (say g(x)) using any of the following approach.

h(x) = f(x) + g(x)h(x)=f(x)+g(x) h(x) = f(x) - g(x)h(x)=f(x)−g(x)

h(x) = \frac{f(x)}{g(x)}h(x)=

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f(x)

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And many more.

The approach or formula to use depends on the question.

In this case, the combined function is:

f(x) = 75+ 10xf(x)=75+10x

The savings function is given as

s(x) = 85s(x)=85

The allowance function is given as:

a(x) = 10(x - 1)a(x)=10(x−1)

The new function that combined his savings and his allowances is calculated as:

f(x) = s(x) + a(x)f(x)=s(x)+a(x)

Substitute values for s(x) and a(x)

f(x) = 85 + 10(x - 1)f(x)=85+10(x−1)

Open bracket

f(x) = 85 + 10x - 10f(x)=85+10x−10

Collect like terms

mark as brainiest

f(x) = 85 - 10+ 10xf(x)=85−10+10x

f(x) = 75+ 10xf(x)=75+10x

4 0
2 years ago
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Read 2 more answers
60 degrees,60 degrees,4cm Do the following conditions make a unique triangle, more than one triangle or no triangle possible?
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NOT NECESSARILY would a triangle be equilateral if one of its angles is 60 degrees. To be an equilateral triangle (a triangle in which all 3 sides have the same length), all 3 angles of the triangle would have to be 60°-angles; however, the triangle could be a 30°-60°-90° right triangle in which the side opposite the 30 degree angle is one-half as long as the hypotenuse, and the length of the side opposite the 60 degree angle is √3/2 as long as the hypotenuse. Another of possibly many examples would be a triangle with angles of 60°, 40°, and 80° which has opposite sides of lengths 2, 1.4845 (rounded to 4 decimal places), and 2.2743 (rounded to 4 decimal places), respectively, the last two of which were determined by using the Law of Sines: "In any triangle ABC, having sides of length a, b, and c, the following relationships are true: a/sin A = b/sin B = c/sin C."¹

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