Answer: It would be 13.0 hours.
Step-by-step explanation: It would be 13.0 hours because $11.00 times 13.0hours is $143.00 so just do 11*13 and you get 143.
Okay here:
Let L be the length and W the width of the rectangle.
1. w=2•L
2. 2L+2W=36
Substitute eq. 1 into eq. 2,
2L+2•(2•L)=36
2L+4L=36
6L=36
L=9
Not done yet, then from eq. 1,
W=2•L
W=2•9
W=18 <------- Your answer. :)
According to the secant-tangent theorem, we have the following expression:
Now, we solve for <em>x</em>.
Then, we use the quadratic formula:
![x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Where a = 1, b = 6, and c = -315.
![\begin{gathered} x_{1,2}=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-315)}}{2\cdot1} \\ x_{1,2}=\frac{-6\pm\sqrt[]{36+1260}}{2}=\frac{-6\pm\sqrt[]{1296}}{2} \\ x_{1,2}=\frac{-6\pm36}{2} \\ x_1=\frac{-6+36}{2}=\frac{30}{2}=15 \\ x_2=\frac{-6-36}{2}=\frac{-42}{2}=-21 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B6%5E2-4%5Ccdot1%5Ccdot%28-315%29%7D%7D%7B2%5Ccdot1%7D%20%5C%5C%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B36%2B1260%7D%7D%7B2%7D%3D%5Cfrac%7B-6%5Cpm%5Csqrt%5B%5D%7B1296%7D%7D%7B2%7D%20%5C%5C%20x_%7B1%2C2%7D%3D%5Cfrac%7B-6%5Cpm36%7D%7B2%7D%20%5C%5C%20x_1%3D%5Cfrac%7B-6%2B36%7D%7B2%7D%3D%5Cfrac%7B30%7D%7B2%7D%3D15%20%5C%5C%20x_2%3D%5Cfrac%7B-6-36%7D%7B2%7D%3D%5Cfrac%7B-42%7D%7B2%7D%3D-21%20%5Cend%7Bgathered%7D)
<h2>Hence, the answer is 15 because lengths can't be negative.</h2>
Answer:
B = (- 4, 
)
Step-by-step explanation:
Using the Section formula
= 
= 
= 
= - 4
= 
= 
= 
=
Thus coordinates of B = (- 4, )
X= 130°
Step-by-step explanation:
From E, draw EF || AB || CD.
Now, EF || CD and CE is the transversal.
So; <DCE + <CEF = 180° [co. int. <s]
x° + <CEF = 180°
<CEF = (180° – x°).
Again, EF || AB and AE is the transversal.
So; <BAE + <AEF = 180° 01 [co. int. <s ]
105° + <AEC+ <CEF = 180°
105° +25° + (180° – x°) = 180°
x° = 130°
I hope I helped you^_^
