Answer:
Step-by-step explanation:
is given value in dollars of a stock in t months after it is purchased.
a) Substitute 1 for t
![V(1) = 42(1-e^{-1.5} )+36 \\=67.852\\=67.85](https://tex.z-dn.net/?f=V%281%29%20%3D%2042%281-e%5E%7B-1.5%7D%20%29%2B36%20%5C%5C%3D67.852%5C%5C%3D67.85)
V(12) = ![V(1) = 42(1-e^{-1.5*12} )+36 \\=77.00](https://tex.z-dn.net/?f=V%281%29%20%3D%2042%281-e%5E%7B-1.5%2A12%7D%20%29%2B36%20%5C%5C%3D77.00)
b) Find derivative for V
![V'(t) = 42(-1.5) e^{-1.5t} )\\\\=-63e^{-1.5t}](https://tex.z-dn.net/?f=V%27%28t%29%20%3D%2042%28-1.5%29%20e%5E%7B-1.5t%7D%20%29%5C%5C%5C%5C%3D-63e%5E%7B-1.5t%7D)
c) When V(t) = 75
![V(t)=75= 42(1-e ^{1.5t} ) +36\\1-e ^{1.5t=0.9286\\=-2.639](https://tex.z-dn.net/?f=V%28t%29%3D75%3D%2042%281-e%20%5E%7B1.5t%7D%20%29%20%2B36%5C%5C1-e%20%5E%7B1.5t%3D0.9286%5C%5C%3D-2.639)
d) As t tends to infinity, exponent being in negative t tends to 0
So V tends to 42(1-0)+36 = 78
Answer:
Total area = 237.09 cm²
Step-by-step explanation:
Given question is incomplete; here is the complete question.
Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)
From the figure attached,
Area of the right triangle I = ![\frac{1}{2}(\text{Base})\times (\text{Height})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Ctext%7BBase%7D%29%5Ctimes%20%28%5Ctext%7BHeight%7D%29)
Area of ΔADC = ![\frac{1}{2}(\text{CD})(\text{AD})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Ctext%7BCD%7D%29%28%5Ctext%7BAD%7D%29)
= ![\frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Csqrt%7B%28AC%29%5E2-%28AD%29%5E2%7D%29%28%5Ctext%7BAD%7D%29)
= ![\frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Csqrt%7B%2813%29%5E2-%2819-7%29%5E2%7D%20%29%2819-7%29)
= ![\frac{1}{2}(\sqrt{169-144})(12)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Csqrt%7B169-144%7D%29%2812%29)
= ![\frac{1}{2}(5)(12)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%285%29%2812%29)
= 30 cm²
Area of equilateral triangle II = ![\frac{\sqrt{3} }{4}(\text{Side})^2](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B4%7D%28%5Ctext%7BSide%7D%29%5E2)
Area of equilateral triangle II = ![\frac{\sqrt{3}}{4}(13)^2](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%2813%29%5E2)
= ![\frac{(1.73)(169)}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%281.73%29%28169%29%7D%7B4%7D)
= 73.0925
≈ 73.09 cm²
Area of rectangle III = Length × width
= CF × CD
= 7 × 5
= 35 cm²
Area of trapezium EFGH = ![\frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28%5Ctext%7BEF%7D%2B%5Ctext%7BGH%7D%29%28%5Ctext%7BFJ%7D%29)
Since, GH = GJ + JK + KH
17 = ![\sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B9%5E%7B2%7D-x%5E%7B2%7D%7D%2B5%2B%5Csqrt%7B%2815%29%5E2-x%5E%7B2%7D%7D)
12 = ![\sqrt{81-x^2}+\sqrt{225-x^2}](https://tex.z-dn.net/?f=%5Csqrt%7B81-x%5E2%7D%2B%5Csqrt%7B225-x%5E2%7D)
144 = (81 - x²) + (225 - x²) + 2![\sqrt{(81-x^2)(225-x^2)}](https://tex.z-dn.net/?f=%5Csqrt%7B%2881-x%5E2%29%28225-x%5E2%29%7D)
144 - 306 = -2x² + ![2\sqrt{(81-x^2)(225-x^2)}](https://tex.z-dn.net/?f=2%5Csqrt%7B%2881-x%5E2%29%28225-x%5E2%29%7D)
-81 = -x² + ![\sqrt{(81-x^2)(225-x^2)}](https://tex.z-dn.net/?f=%5Csqrt%7B%2881-x%5E2%29%28225-x%5E2%29%7D)
(x² - 81)² = (81 - x²)(225 - x²)
x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴
144x² - 11664 = 0
x² = 81
x = 9 cm
Now area of plot IV = ![\frac{1}{2}(5+17)(9)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%285%2B17%29%289%29)
= 99 cm²
Total Area of the land = 30 + 73.09 + 35 + 99
= 237.09 cm²
Um, here is the problem. WE NEED A PICTURE TO ACTUALLY TELL YOU THE ANSWER! DO YOU WANT AN ANSWER OR NOT?????
3.50x + 5 < 30
Ellie wants to spend less than 30, so it’ll be less than.
Answer:
![x + 2y= 2](https://tex.z-dn.net/?f=x%20%2B%202y%3D%202)
Step-by-step explanation:
Given
Points:
![F = (4,9)](https://tex.z-dn.net/?f=F%20%3D%20%284%2C9%29)
![G = (1,3)](https://tex.z-dn.net/?f=G%20%3D%20%281%2C3%29)
Required
Determine the equation of line that is perpendicular to the given points and that pass through ![(2,0)](https://tex.z-dn.net/?f=%282%2C0%29)
First, we need to determine the slope, m of FG
![m = \frac{y_2 - y_1}{x_2 - x_1}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7By_2%20-%20y_1%7D%7Bx_2%20-%20x_1%7D)
Where
--- ![(x_1,y_1)](https://tex.z-dn.net/?f=%28x_1%2Cy_1%29)
--- ![(x_2,y_2)](https://tex.z-dn.net/?f=%28x_2%2Cy_2%29)
![m = \frac{3 - 9}{1 - 4}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B3%20-%209%7D%7B1%20-%204%7D)
![m = \frac{- 6}{- 3}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B-%206%7D%7B-%203%7D)
![m =2](https://tex.z-dn.net/?f=m%20%3D2)
The question says the line is perpendicular to FG.
Next, we determine the slope (m2) of the perpendicular line using:
![m_2 = -\frac{1}{m}](https://tex.z-dn.net/?f=m_2%20%3D%20-%5Cfrac%7B1%7D%7Bm%7D)
![m_2 = -\frac{1}{2}](https://tex.z-dn.net/?f=m_2%20%3D%20-%5Cfrac%7B1%7D%7B2%7D)
The equation of the line is then calculated as:
![y - y_1 = m_2(x - x_1)](https://tex.z-dn.net/?f=y%20-%20y_1%20%3D%20m_2%28x%20-%20x_1%29)
Where
![m_2 = -\frac{1}{2}](https://tex.z-dn.net/?f=m_2%20%3D%20-%5Cfrac%7B1%7D%7B2%7D)
![(x_1,y_1) = (2,0)](https://tex.z-dn.net/?f=%28x_1%2Cy_1%29%20%3D%20%282%2C0%29)
![y - 0 = -\frac{1}{2}(x - 2)](https://tex.z-dn.net/?f=y%20-%200%20%3D%20-%5Cfrac%7B1%7D%7B2%7D%28x%20-%202%29)
![y = -\frac{1}{2}(x - 2)](https://tex.z-dn.net/?f=y%20%20%3D%20-%5Cfrac%7B1%7D%7B2%7D%28x%20-%202%29)
![y = -\frac{1}{2}x + 1](https://tex.z-dn.net/?f=y%20%20%3D%20-%5Cfrac%7B1%7D%7B2%7Dx%20%2B%201)
Multiply through by 2
![2y = -x + 2](https://tex.z-dn.net/?f=2y%20%3D%20-x%20%2B%202)
Add x to both sides
![x + 2y= -x +x+ 2](https://tex.z-dn.net/?f=x%20%2B%202y%3D%20-x%20%2Bx%2B%202)
![x + 2y= 2](https://tex.z-dn.net/?f=x%20%2B%202y%3D%202)
Hence, the line of the equation is ![x + 2y= 2](https://tex.z-dn.net/?f=x%20%2B%202y%3D%202)