Find the area of the figure below:

Mathematics

1 answer:

SpyIntel [72]2 years ago

3 0

The total area is 407.4262m²

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How to solve for area of triangle where A=67 degrees, b=11yd, and c=24yd

skad [1K]

Use the formula of an area of a tringle:

$A=\dfrac{1}{2}bc\sin A$

We have b = 11 yd, c = 24 yd and m∠A = 67°

$A=\dfrac{1}{2}(11)(24)\sin67^o$

$\sin67^o\approx0.9205\\\A=(11)(12)(0.9205)=121.506\ yd^2$

4 0

3 years ago

Use the distributive property. 4(5x + 3y - 7) = 20x + 12y + [?]

katen-ka-za [31]

Answer:

- 28

Step-by-step explanation:

4(5x + 3y - 7) ← multiply each term in the parenthesis by 4

= 20x + 12y - 28

6 0

2 years ago

Read 2 more answers

The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

Firdavs [7]

Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.

This means the area is given by the integral,

$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.