Please help with how to find the area of the shape with work

10 squares.

Step-by-step explanation:

We can do it with integrals.

If we take the bottom vertex as the point (0,0), the top right vertex as the point (0,4) and the left vertex as the point (5, 6)

then the area of the triangle is the area enclosed by two lines between the values x = 0 and x = 5.

the two lines are:

the bottom is the one that passes through (0,0) and (5,6)

f(x) = y = s*x (because it passes through the point (0,0))

and the slope is s = 6/5.

so f(x) =  y = (6/5)*x.

The top line is the one that passes through (0,4) and (5,6)

the y-intercept is b = 4, and the slope is:

s = (6 - 4)/(5 - 0) = 2/5.

g(x) = y = (2/5)*x + 4.

Now, the area enclosed for the triangle is equal to:

this is equal to:

= (1/2)(-4/5)*(5)^2 + 4*5 - 0 = 10

The area is 10 squares

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation: