I think its "Yes it can because -3.5 lies to the left of -1.5"
Answer:
a. x=2, y=13
b. x=14, y=25
c. x=−4, y=13
Step-by-step explanation:
Answer:
The answer is 44 months.
Step-by-step explanation:12*5=60
60 months - 16 months = 44
Answer:
4.5 sq. units.
Step-by-step explanation:
The given curve is ![y = (3x)^{\frac{1}{2} }](https://tex.z-dn.net/?f=y%20%3D%20%283x%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D)
⇒
...... (1)
This curve passes through (0,0) point.
Now, the straight line is y = 3x - 6 ....... (2)
Now, solving (1) and (2) we get,
![y^{2} - y - 6 = 0](https://tex.z-dn.net/?f=y%5E%7B2%7D%20-%20y%20-%206%20%3D%200)
⇒ (y - 3)(y + 2) = 0
⇒ y = 3 or y = -2
We will consider y = 3.
Now, y = 3x - 6 has zero at x = 2.
Therefor, the required are = ![\int\limits^3_0 {(3x)^{\frac{1}{2} } } \, dx - \int\limits^3_2 {(3x - 6)} \, dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5E3_0%20%7B%283x%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20%7D%20%5C%2C%20dx%20-%20%5Cint%5Climits%5E3_2%20%7B%283x%20-%206%29%7D%20%5C%2C%20dx)
= ![\sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20%5B%7B%5Cfrac%7Bx%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%7D%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%7D%5D%5E%7B3%7D%20_%7B0%7D%20-%20%5B%5Cfrac%7B3x%5E%7B2%7D%20%7D%7B2%7D%20-%206x%20%5D%5E%7B3%7D%20_%7B2%7D)
= ![[\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} } }{3}] - [13.5 - 18 - 6 + 12]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%5Csqrt%7B3%7D%5Ctimes%202%20%5Ctimes%203%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%7D%7B3%7D%5D%20-%20%5B13.5%20-%2018%20-%206%20%2B%2012%5D)
= 6 - 1.5
= 4.5 sq. units. (Answer)