Answer:
∠ WZX = 50°
XW is not an altitude.
Step-by-step explanation:
16. See the attached figure.
XW is the angle bisector of ∠ YXZ, hence, ∠ WXY = ∠ WXZ
Now, given that ∠ YXZ = 8x + 34 and ∠ WXY = 10x - 13
Hence, ∠ YXZ = 2 ∠ WXY
⇒ 8x + 34 = 2(10x - 13)
⇒ 8x + 34 = 20x - 26
⇒ 12x = 60
⇒ x = 5.
Hence, ∠ XZY = ∠ WZX = 10x = 50° (Answer)
Now, ∠ WXZ = ∠ WXY = 10x - 13 = 37°
Hence, from Δ WXZ,
∠ WZX + ∠ WXZ + ∠ XWZ = 180°
⇒ 50° + 37° + ∠ XWZ = 180°
⇒ ∠ XWZ = 93° ≠ 90°
Hence, XW is not an altitude. (Answer)
Answer:
Step-by-step explanation:
<u>Given expression:</u>
<u>Find its value when x = 4 and y = 1, substitute:</u>
- 3(4 + 1) + 2*4² + (1 + 1)² =
- 3*5 + 2*16 + 2² =
- 15 + 32 + 4 =
- 51
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
__
You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
__
Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
Answer: 2 1/2
Step-by-step explanation: Divide using long division. The whole number portion will be the number of times the denominator of the original fraction divides evenly into the numerator of the original fraction, and the fraction portion of the mixed number will be the remainder of the original fraction division over the denominator of the original fraction. Hope this helps!! PLZ mark me!!!
Answer:
(A) Set A is linearly independent and spans 
. Set is a basis for .
Step-by-Step Explanation
<u>Definition (Linear Independence)</u>
A set of vectors is said to be linearly independent if at least one of the vectors can be written as a linear combination of the others. The identity matrix is linearly independent.
<u>Definition (Span of a Set of Vectors)</u>
The Span of a set of vectors is the set of all linear combinations of the vectors.
<u>Definition (A Basis of a Subspace).</u>
A subset B of a vector space V is called a basis if: (1)B is linearly independent, and; (2) B is a spanning set of V.
Given the set of vectors ![A= \left(\begin{array}{[c][c][c][c]}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\end{array} \right)](https://tex.z-dn.net/?f=A%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7B%5Bc%5D%5Bc%5D%5Bc%5D%5Bc%5D%7D1%20%26%200%20%26%200%20%26%200%5C%5C%200%20%26%201%20%26%200%20%26%201%5C%5C%200%20%26%200%20%26%201%20%26%201%5Cend%7Barray%7D%20%5Cright%29%20)
, we are to decide which of the given statements is true:
In Matrix ![A= \left(\begin{array}{[c][c][c][c]}(1) & 0 & 0 & 0\\ 0 & (1) & 0 & 1\\ 0 & 0 & (1) & 1\end{array} \right)](https://tex.z-dn.net/?f=A%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7B%5Bc%5D%5Bc%5D%5Bc%5D%5Bc%5D%7D%281%29%20%26%200%20%26%200%20%26%200%5C%5C%200%20%26%20%281%29%20%26%200%20%26%201%5C%5C%200%20%26%200%20%26%20%281%29%20%26%201%5Cend%7Barray%7D%20%5Cright%29%20)
, the circled numbers are the pivots. There are 3 pivots in this case. By the theorem that The Row Rank=Column Rank of a Matrix, the column rank of A is 3. Thus there are 3 linearly independent columns of A and one linearly dependent column. has a dimension of 3, thus any 3 linearly independent vectors will span it. We conclude thus that the columns of A spans .
Therefore Set A is linearly independent and spans . Thus it is basis for .
