Answer:
15 square units.
Step-by-step explanation:
To find the area, we must find the width and length of the rectangle. To find the width and length, we need to find the length of AB and the length of BC. We use the distance formula.
AB: A(-8, 3); B(-3, 3).

= 
= 
= 
= 5
BC: B(-3, 3); C(-3, 6).

= 
= 
= 
= 
= 3
Now that we have both the width and the length, we can solve for the area using A = lw, where A is the area, l is the length, and w is the width. In this case, l = 5 and w = 3.
A = lw
A = 5 * 3
A = 15
So, the area of the rectangle is 15 square units.
Hope this helps!
Answer:
∫ₑ°° 1 / (x (ln x)¹⁰) dx
∫₁°° x⁻³ dx
Step-by-step explanation:
A p-series 1 / xᵖ converges if p > 1.
∫ₑ°° 1 / (x (ln x)¹⁰) dx
If u = ln x, then du = 1/x dx.
When x = e, u = 1. When x = ∞, u = ∞.
= ∫₁°° 1 / (u¹⁰) du
p = 10, converges
∫₁₀°° x^(-⅔) dx
= ∫₁₀°° 1 / (x^⅔) dx
p = ⅔, diverges
∫₁°° 2 / x^0.5 dx
= 2 ∫₁°° 1 / x^0.5 dx
p = 0.5, diverges
∫₁°° x⁻³ dx
= ∫₁°° 1 / x³ dx
p = 3, converges
∫₂°° 1/(3x) dx
= ⅓ ∫₂°° 1/x dx
p = 1, diverges
<span>Draw a graph. Label the x- and y- axis. Choose a <span>range </span>that includes the maximums and minimums from the given data.</span><span> For example, our x-values go from 3 to 6.3, so a range from 3 to 7 would be approprite.
</span>
Let a = any number
So, 1 times a = a.
In other words, any number times 1 = any number.
Two samples:
1. 45 x 1 = 45
2. -567 x 1 = -567
Understand?
4x = 8, so x = 2
5y + 4 = 4, so 5y = 0, so y = 0
(2,0)