We know that
the scale is <span>is 1/2 inch = 4 feet
using a graph tool
</span><span>the approximate measurements in the drawing are
</span>5 cm x 4 cm
convert to in
5/2.54--------> 1.97 in
4/2.54------> 1.57 in
convert drawing to actual
if 0.5 in---------> 4 ft
1.97in-------> X
X=1.97*4/0.5------> X=15.76 ft
if 0.5 in---------> 4 ft
1.57 in-------> X
X=1.57*4/0.5------> X=12.56 ft
<span>the approximate area of Quinto's bedroom is 12.56*15.76=197.95 ft</span>²
the answer is
the option C)192 ft²
Answer:
B: 3
Step-by-step explanation:
The expression would be 14m + 23 = 65.
14m + 23 - 23 = 65 - 23 =
14m = 42
14m ÷ 14 = 42 ÷ 14 =
m = 3
I hope this helped and if it did I would appreciate it if you marked me Brainliest. Thank you and have a nice day!
Answer:
The solution to this question can be defined as follows:
Step-by-step explanation:
Please find the complete question in the attached file.
![A = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%5Cend%7Barray%7D%5Cright%5D)
now for given values:
![\left[\begin{array}{ccc} \frac{3}{4} - \lambda & \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2} - \lambda & 0\\ -\frac{1}{4}& -\frac{1}{4} & 0 -\lambda \end{array}\right]=0 \\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%20-%5Clambda%20%5Cend%7Barray%7D%5Cright%5D%3D0%20%5C%5C%5C%5C)
![\to (\frac{3}{4} - \lambda ) [-\lambda (\frac{1}{2} - \lambda ) -0] - 0 - \frac{1}{4}[0- \frac{1}{2} (\frac{1}{2} - \lambda )] =0 \\\\\to (\frac{3}{4} - \lambda ) [(\frac{\lambda}{2} + \lambda^2 )] - \frac{1}{4}[\frac{\lambda}{2} - \frac{1}{4}] =0 \\\\\to (\frac{3}{8}\lambda + \frac{3}{4} \lambda^2 - \frac{\lambda^2}{2} - \lambda^3 - \frac{\lambda}{8} + \frac{1}{16}=0 \\\\\to (\lambda - \frac{1}{2}) (\lambda -\frac{1}{4}) (\lambda - \frac{1}{2}) =0\\\\](https://tex.z-dn.net/?f=%5Cto%20%20%28%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%29%20%5B-%5Clambda%20%28%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%29%20-0%5D%20-%200%20-%20%5Cfrac%7B1%7D%7B4%7D%5B0-%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%29%5D%20%3D0%20%5C%5C%5C%5C%5Cto%20%20%28%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%29%20%5B%28%5Cfrac%7B%5Clambda%7D%7B2%7D%20%2B%20%5Clambda%5E2%20%29%5D%20-%20%5Cfrac%7B1%7D%7B4%7D%5B%5Cfrac%7B%5Clambda%7D%7B2%7D%20-%20%20%5Cfrac%7B1%7D%7B4%7D%5D%20%3D0%20%5C%5C%5C%5C%5Cto%20%20%28%5Cfrac%7B3%7D%7B8%7D%5Clambda%20%2B%20%5Cfrac%7B3%7D%7B4%7D%20%5Clambda%5E2%20-%20%5Cfrac%7B%5Clambda%5E2%7D%7B2%7D%20-%20%5Clambda%5E3%20-%20%5Cfrac%7B%5Clambda%7D%7B8%7D%20%2B%20%5Cfrac%7B1%7D%7B16%7D%3D0%20%5C%5C%5C%5C%5Cto%20%28%5Clambda%20-%20%5Cfrac%7B1%7D%7B2%7D%29%20%28%5Clambda%20-%5Cfrac%7B1%7D%7B4%7D%29%20%28%5Clambda%20-%20%5Cfrac%7B1%7D%7B2%7D%29%20%3D0%5C%5C%5C%5C)
In point b:
Its
spectral radius is less than 1 hence matrix is convergent.
In point c:
![\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) = \left(\begin{array}{c}3&1&2\end{array}\right) , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\ \to x^{(k+1)} = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right] \\\\](https://tex.z-dn.net/?f=%5Cto%20c%5E%7B%28k%2B1%29%7D%20%3D%20A%20x%5E%7Bk%7D%2BC%20%5C%5C%5C%5C%5Cto%20x%280%29%20%3D%20%20%20%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7D3%261%262%5Cend%7Barray%7D%5Cright%29%20%20%2C%20c%20%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7D2%262%264%5Cend%7Barray%7D%5Cright%29%5C%5C%5C%5C%20%20%5Cto%20x%5E%7B%28k%2B1%29%7D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%5Cend%7Barray%7D%5Cright%5D%20x%5Ek%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%262%264%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%5C%5C)
after solving the value the answer is
:
![\lim_{k \to \infty} x^k=o = \left[\begin{array}{c}0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Clim_%7Bk%20%5Cto%20%5Cinfty%7D%20x%5Ek%3Do%20%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%260%260%5Cend%7Barray%7D%5Cright%5D)
SIN(θ) = OPPOSITE / HYPOTENUSE = X / Y
COS(θ) = ADJACENT / HYPOTENUSE = 2 / Y
TAN(θ) = OPPOSITE / ADJACENT = X / 2
CSC(θ) = 1 / SIN(θ) = 1 / (X / Y) = Y / X
SEC(θ) = 1 / COS (θ) = 1 / (2 / Y) = Y / 2
COT (θ) = 1 / TAN(θ) = 1 / (X / 2) = 2 / X
HUNEY, YOU HAVE ALL OF THE VALUES, NOW JUST SOLVE FOR X AND Y THEN...
FOR EXAMPLE:
SIN(θ) = X / Y X = Y * SIN(θ)
COS(θ) = 2 / Y Y = 2 / COS(θ)
TAN(θ) = X / 2 X = 2 * TAN(θ)
SO ON, SO FORTH HUN U GET IT
Answer:
Expressions A, E, and F
Step-by-step explanation:
In general, when negative integers are multiplied; if the negative integers are even, then the product would be positive. But if the negative integers are odd, then the product would be negative.
Thus,
A. (−5)(−0.4)(1.8) = (2.0)(1.8)
= 3.6
B. (-45)(-19)(-112) = (855)(-112)
= -95760
C. −1(5+17) = -1(22)
= -22
D. (−5)(−0.4)(1.8)(−3.25) = (2.0)(-5.85)
= -11.7
E. (-49)(-34)(-12)(-12) = (1666)(144)
= 239904
F. −1(3.5−7) = -1(-3.5)
= 3.5
Thus expressions A, E and F have positive products.
