Answer:
<h3>B. It has infinite solutions</h3>
Step-by-step explanation:
Given the system of equations:
2t + w = 10 ..... 1
4t = 20 − 2w ... 2
From 1:
w = 10-2t ...3
Substitute 3 into 2 to have;
4t = 20 - 2(10-2t)
4t = 20-20+4t
4t = 4t
Let t = k
Substitute t = k into 1 and get w;
From 1: 2t + w = 10
2k + w =10
w = 10 - 2k
<em>k can take any integers. This shows that the solution to the equation is infinite</em>
<em></em>
Since LM = AM, point M must be on the perpendicular bisector of AL. Since AM = BM, BL must be perpendicular to AL. This makes ∆ALC a right triangle with hypotenuse AC twice the length of side AL. Hence ∠LAC = ∠LAB = 60°, and AL is angle bisector, median, and altitude.
ΔABC is isosceles with ∠A = 120°, and ∠B = ∠C = 30°.
Answer:
wrong as his calculation was incorrect at initial level itself when he found the result of the division
Step-by-step explanation:
Andre said
3 ÷ ![\frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D)
![=3 \times \frac{3}{2}](https://tex.z-dn.net/?f=%3D3%20%5Ctimes%20%5Cfrac%7B3%7D%7B2%7D)
![=\frac{3 \times 3}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B3%20%5Ctimes%203%7D%7B2%7D)
![=\frac{9}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B9%7D%7B2%7D)
but andre calculated it as
![=4\tfrac{1}{3}](https://tex.z-dn.net/?f=%3D4%5Ctfrac%7B1%7D%7B3%7D)
![=\frac{4 \times 3 + 1}{3}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B4%20%5Ctimes%203%20%2B%201%7D%7B3%7D)
![=\frac{13}{3}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B13%7D%7B3%7D)
Hence his calculation was incorrect at initial level itself
Answer:
0.7 minute
Step-by-step explanation:
Given that each (1) minute a light use 9.4 watts.
So If the light bulb used a total of 6.58 watts, the minutes was on it is:
=
= 0.7 minute
So the the minutes was on it is 0.7
Hope it will find you well!
I think you have to pay attention in class for this