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3 years ago

Find the area of the following shape. You must show all work to receive credit A (-4, 0), B(0, 4), C(2, 2), D(1,0)

Mathematics

1 answer:

natulia [17]3 years ago

7 0

9514 1404 393

Answer:

13 square units

Step-by-step explanation:

See the attached figure.

The horizontal dashed line divides the figure into two pieces:

a triangle 4 wide and 2 high
a trapezoid 2 high with bases 5 and 4

The appropriate area formulas can be used to find the areas of these. The figure's area will be their sum.

A = (1/2)bh = (1/2)(4)(2) = 4 . . . . square units

A = (1/2)(b1 +b2)h = (1/2)(5 +4)(2) = 9 . . . . square units

The area of the shape is ...

(4 +9) square units = 13 square units

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Answer:

Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 0

3 years ago

I need help with this

Triss [41]

Since 12 is hypotenuse, formula we would use is:
12^2=5^2+x^2
144=25+x^2
x^2= 144-25
x^2=119
x= √119
x=10.9087
Rounded to nearest tenth,
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4 0

3 years ago

The function f has a domain of {1, 3, 5, 7} and a range of {2, 4, 6}. Could f be represented by {(1, 2), (3, 4), (5, 6), (7, 2)}

bezimeni [28]

Answer:

Yes

Step-by-step explanation:

- Yes, a function with those domain and range values could certainly be represented by those points.

- (1,2) and (7,2) are points with separate domain values but identical range values: this is allowed because every x value is only assigned one y value (remember, one number x in domain cannot have two different numbers y in the range).

- Another way is the use of Vertical Line test of the curve passing through to the points {(1, 2), (3, 4), (5, 6), (7, 2)}. sketch down multiple vertical lines passing through the curve. If any particular vertical line intersects the curve at more than one point on the curve then the curve is not a function of y = f(x).

- You would see that all vertical lines are free this invalidation. Hence, y is a function of x having points {(1, 2), (3, 4), (5, 6), (7, 2)}.

5 0

4 years ago