Answer:
38m and 90cm
1m = 100cm
3890cm = 38m and 90cm
Step-by-step explanation:
1m = 100cm
3890cm = 38m and 90cm
Answer:
The length would be 1.5x + 5.
Step-by-step explanation:
So first with this, note that 2 length + 2 height gives the perimeter of the rectangle.
Given this:
2(1/2x + 4) + 2(?) = 4x+18
x+8 + 2(?) = 4x+18
2(?) = 3x+10
? = 1.5x+5
1.5x + 5 = 1.5x + 5
The length would be 1.5x + 5.
To double check, by plugging in 1.5x + 5 back into the equation for the perimeter, you will get the same perimeter:
2(1.5x + 5) + 2(1/2x+4) = P
3x+10 + x+8 = P
4x + 18 = P
Answer:
Increasing on it's domain
because the slope is positive.
The domain and range are both all real numbers, also known as
.
Step-by-step explanation:
All domain really means is what numbers can you plug in and you get number back from your function.
I should be able to plug in any number into 3x+2 and result in a number. There are no restrictions for x on 3x+2.
The domain is all real numbers.
In interval notation that is
.
Now the range is the set of numbers that get hit by y=3x+2.
Well y=3x+2 is a linear function that is increasing. I know it is increasing because the slope is positive 3. I wrote out the positive part because that is the item you focus on in a linear equation to determine if is increasing or decreasing.
If slope is positive, then the line is increasing.
If slope is negative, then the line is decreasing.
So y=3x+2 hits all values of y because it is increasing forever. The range is all real numbers. In interval notation that is
.
Answer:
0
Step-by-step explanation:
∫∫8xydA
converting to polar coordinates, x = rcosθ and y = rsinθ and dA = rdrdθ.
So,
∫∫8xydA = ∫∫8(rcosθ)(rsinθ)rdrdθ = ∫∫8r²(cosθsinθ)rdrdθ = ∫∫8r³(cosθsinθ)drdθ
So we integrate r from 0 to 9 and θ from 0 to 2π.
∫∫8r³(cosθsinθ)drdθ = 8∫[∫r³dr](cosθsinθ)dθ
= 8∫[r⁴/4]₀⁹(cosθsinθ)dθ
= 8∫[9⁴/4 - 0⁴/4](cosθsinθ)dθ
= 8[6561/4]∫(cosθsinθ)dθ
= 13122∫(cosθsinθ)dθ
Since sin2θ = 2sinθcosθ, sinθcosθ = (sin2θ)/2
Substituting this we have
13122∫(cosθsinθ)dθ = 13122∫(1/2)(sin2θ)dθ
= 13122/2[-cos2θ]/2 from 0 to 2π
13122/2[-cos2θ]/2 = 13122/4[-cos2(2π) - cos2(0)]
= -13122/4[cos4π - cos(0)]
= -13122/4[1 - 1]
= -13122/4 × 0
= 0