To find the product of (4x-5y)^2,
we can rewrite the problem as:
(4x-5y)(4x-5y) (two times because it is squared)
Now, time to use that old method we learned in middle school:
FOIL. (Firsts, Outers, Inners, and Lasts)
FOIL can help us greatly in this scenario.
Let's start by multiplying the 'Firsts' together:
4x * 4x = <em>16x^2</em>
Now, lets to the 'Outers':
4x * -5y = <em>-20xy</em>
Next, we can multiply the 'Inners':
-5y * 4x = <em>-20xy</em>
Finally, let's do the 'Lasts':
-5y * -5y = <em>25y</em>^2
Now, we can take the products of these equations from FOIL and combine like terms. We have: 16x^2, -20xy, -20xy, and 25y^2.
-20xy and -20xy make -40xy.
The final equation (product of (4x-5y)^2) is:
16x^2 - 40xy + 25y^2
Hope I helped! If any of my math is wrong, please report and let me know!
Have a good one.
Answer:
12 cm
Step-by-step explanation:
The formula for the area of a trapezoid is written as:
1/2(b1 + b2)h
h = height = 16 cm
b1 = Length of one parallel side = 9cm
b2 = Length of second parallel side = ?
Area of trapezoid = 168cm²
The formula to find the length of the second parallel side =
b2 = 2A/h - b1
b2 = 2 × 168/16 - 9
b2 = 336/16 - 9
b2 = 21 - 9
b2 = 12cm
Therefore, the length of the second parallel side is 12 cm
Between 0 and 1 because it goes 0, 1/10, 2/10, and so on and so forth till 1
Answer:
100
Step-by-step explanation:
5x4=20
20x5=100
The purpose of the tensor-on-tensor regression, which we examine, is to relate tensor responses to tensor covariates with a low Tucker rank parameter tensor/matrix without being aware of its intrinsic rank beforehand.
By examining the impact of rank over-parameterization, we suggest the Riemannian Gradient Descent (RGD) and Riemannian Gauss-Newton (RGN) methods to address the problem of unknown rank. By demonstrating that RGD and RGN, respectively, converge linearly and quadratically to a statistically optimal estimate in both rank correctly-parameterized and over-parameterized scenarios, we offer the first convergence guarantee for the generic tensor-on-tensor regression. According to our theory, Riemannian optimization techniques automatically adjust to over-parameterization without requiring implementation changes.
Learn more about tensor-on-tensor here
brainly.com/question/16382372
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