Answer:
When the inputs are 1 and 0, the output is zero.
Answer:
Statements Reasons
1. 2x + 11 = 15 1. Given
2. 2x = 4 2. Subtraction Property of Equality
3. X = 2 3. Division Property of Equality
Step-by-step explanation:
An equation can be solved and its solution proven using algebraic theorems and properties. To create a proof, form two columns. Label one side Statements and the other Reasons.
Begin your proof listing the any information given to you. List as the reason - Given.
Then list the next step which here would be to subtract by 11 on both side. The reason is Subtraction Property of Equality. Subtraction is the inverse of addition. Inverse axiom is another acceptable reason.
Then divide both sides by 2. The reason is Division Property of Equality or Inverse axiom once again. See the proof below.
Statements Reasons
1. 2x + 11 = 15 1. Given
2. 2x = 4 2. Subtraction Property of Equality
3. X = 2 3. Division Property of Equality
what's up? the answer to this is 339+340+341
best of luck with your studies
Answer with Step-by-step explanation:
Suppose that a matrix has two inverses B and C
It is given that AB=I and AC=I
We have to prove that Inverse of matrix is unique
It means B=C
We know that
B=BI where I is identity matrix of any order in which number of rows is equal to number of columns of matrix B.
B=B(AC)
B=(BA)C
Using associative property of matrix
A (BC)=(AB)C
B=IC
Using BA=I
We know that C=IC
Therefore, B=C
Hence, Matrix A has unique inverse .
Answer:
(8√2) / 15
Step-by-step explanation:
A curve bounded by the y-axis is represented by in terms of dy;
When the curve crosses the y-axis, x will be 0. In this case x is the function of t, so we have to solve for x(t) = 0;
0 = t^2 + 2t --- (1)
Solution(s) => t = 0, t = 2
dy = (1/2 * 1/√t)dt --- (2)
Our solutions (0, 2) are our limits. The area of the curve is in the form 
, so now let's introduce the limits of integration, x(t) and dy/dt. Remember, dy/dt = (1/2 * 1/√t) (second equation). 1/2 * 1/√t can be rewritten as 1/2 * t^(-1/2)....
![A\:=\:\int _2^0\:\left(t^2-2t\right)\left(\frac{1}{2}t^{-\frac{1}{2}}\right)dt\\\\= \int _2^0\:\left(\frac{1}{2}t^{\frac{3}{2}}-t^{\frac{1}{2}}\right)dt\\\\= \left[\frac{t^{\frac{5}{2}}}{5}-\frac{2t^{\frac{3}{2}}}{3}\right]_2^0\\\\= 0\:-\:\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\\\\= \frac{8\sqrt{2}}{15}](https://tex.z-dn.net/?f=A%5C%3A%3D%5C%3A%5Cint%20_2%5E0%5C%3A%5Cleft%28t%5E2-2t%5Cright%29%5Cleft%28%5Cfrac%7B1%7D%7B2%7Dt%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5Cright%29dt%5C%5C%5C%5C%3D%20%5Cint%20_2%5E0%5C%3A%5Cleft%28%5Cfrac%7B1%7D%7B2%7Dt%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D-t%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5Cright%29dt%5C%5C%5C%5C%3D%20%5Cleft%5B%5Cfrac%7Bt%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%7D%7B5%7D-%5Cfrac%7B2t%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B3%7D%5Cright%5D_2%5E0%5C%5C%5C%5C%3D%200%5C%3A-%5C%3A%5Cleft%28%5Cfrac%7B4%5Csqrt%7B2%7D%7D%7B5%7D-%5Cfrac%7B4%5Csqrt%7B2%7D%7D%7B3%7D%5Cright%29%5C%5C%5C%5C%3D%20%5Cfrac%7B8%5Csqrt%7B2%7D%7D%7B15%7D)
Your solution is 8√2 / 15
