Answer:
Area of composite figure = 216 cm²
Hence, option A is correct.
Step-by-step explanation:
The composite figure consists of two figures.
1) Rectangle
2) Right-angled Triangle
We need to determine the area of the composite figure, so we need to find the area of an individual figure.
Determining the area of the rectangle:
Given
Length l = 14 cm
Width w = 12 cm
Using the formula to determine the area of the rectangle:
A = wl
substituting l = 14 and w = 12
A = (12)(14)
A = 168 cm²
Determining the area of the right-triangle:
Given
Base b = 8 cm
Height h = 12 cm
Using the formula to determine the area of the right-triangle:
A = 1/2 × b × h
A = 1/2 × 8 × 12
A = 4 × 12
A = 48 cm²
Thus, the area of the figure is:
Area of composite figure = Rectangle Area + Right-triangle Area
= 168 cm² + 48 cm²
= 216 cm²
Therefore,
Area of composite figure = 216 cm²
Hence, option A is correct.
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First, apply this following rule:
÷
×
×
Second, apply this following rule:
×
Third, multiply 1 × 6 to get 6 and 6 × 2 to get 12.
Fourth, find the GCF of 6 and 12.
Factors of 6: 1, 2, 3, 6
Factors of 12: 1, 2, 3, 4, 6, 12
The GCF is 6.
Fifth, divide the numerator by the GCF.
÷
Sixth, divide the denominator by the GCF.
÷
Seventh, collect the new numerator and new denominator.
Answer as fraction:
Answer as decimal: 0.5
Answer:
see explanation
Step-by-step explanation:
so, when a=-3, c is -20. now, to find b when a is -3, plug a (which is -3) into the equation b=4a-5 to get b=-17. so b=-17, c=-20. b is obviously larger (remember the negative sign)
when c is 21, a is 5. plug numbers in to get b=15.
when a is 6, there is no clear value for c, so i'd guess undefined. b however is 19
we can plot the line and the points, for problem 4 and all points above the line means that c is greater than b, and when below it's less than. doing this we get when a is 12, when a is 0, and when a is 5. hope this was clear
a. Assume a is even, so a = 2k for some integer k. Now let a and b be integers such that a divides b and a + b is odd.
Since a divides b, b = an for integer n, and in turn b = 2nk, which means <u>b is even</u> and hence a + b is also even. But this contradicts our initial assumption, so a must be odd.
b. Let n be even, so that n = 2k for some integer k. Then
n² = (2k)² = 4k²
so that n² ≡ 0 (mod 4).
Now let n be odd, so n = 2k + 1 for integer k. Then
n² = (2k + 1)² = 4k² + 4k + 1
so that n² ≡ 1 (mod 4).
Therefore n² is never congruent to 2 (mod 4).