Answer:
D
Step-by-step explanation:
so by process of elimination it can’t be A as it is a 7hr journey
B is also incorrect as she travels 50mph not an additional 50miles per hour (50m in hour one then 100m in hour 2 )
C is also incorrect as she travels 50 miles in ann hour not in 6
D is correct as she has travelled 200 miles
Answer:
Below
I hope its not too complicated
![x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5Ctext%7BW%7D_0%5Cleft%28%5Cfrac%7B%5Cln%20%5Cleft%285%5Cright%29%7D%7B2e%5E%7B%5Cfrac%7B3%5Cln%20%5Cleft%285%5Cright%29%7D%7B2%7D%7D%7D%5Cright%29%7D%7B%5Cln%20%5Cleft%285%5Cright%29%7D%2B%5Cfrac%7B3%7D%7B2%7D)
Step-by-step explanation:
![5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}](https://tex.z-dn.net/?f=5%5E%7B%5Cleft%28-x%5Cright%29%7D%2B7%3D2x%2B4%5C%5C%5C%5C%5Cmathrm%7BPrepare%7D%5C%3A5%5E%7B%5Cleft%28-x%5Cright%29%7D%2B7%3D2x%2B4%5C%3A%5Cmathrm%7Bfor%5C%3ALambert%5C%3Aform%7D%3A%5Cquad%201%3D%5Cleft%282x-3%5Cright%29e%5E%7B%5Cln%20%5Cleft%285%5Cright%29x%7D%5C%5C%5C%5C%5Cmathrm%7BRewrite%5C%3Athe%5C%3Aequation%5C%3Awith%5C%3A%7D%5C%5C%5Cleft%28x-%5Cfrac%7B3%7D%7B2%7D%5Cright%29%5Cln%20%5Cleft%285%5Cright%29%3Du%5Cmathrm%7B%5C%3Aand%5C%3A%7Dx%3D%5Cfrac%7Bu%7D%7B%5Cln%20%5Cleft%285%5Cright%29%7D%2B%5Cfrac%7B3%7D%7B2%7D%5C%5C%5C%5C1%3D%5Cleft%282%5Cleft%28%5Cfrac%7Bu%7D%7B%5Cln%20%5Cleft%285%5Cright%29%7D%2B%5Cfrac%7B3%7D%7B2%7D%5Cright%29-3%5Cright%29e%5E%7B%5Cln%20%5Cleft%285%5Cright%29%5Cleft%28%5Cfrac%7Bu%7D%7B%5Cln%20%5Cleft%285%5Cright%29%7D%2B%5Cfrac%7B3%7D%7B2%7D%5Cright%29%7D)
![Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}](https://tex.z-dn.net/?f=Simplify%5C%5C%5C%5C%5Cmathrm%7BRewrite%7D%5C%3A1%3D%5Cfrac%7B2e%5E%7Bu%2B%5Cfrac%7B3%7D%7B2%7D%5Cln%20%5Cleft%285%5Cright%29%7Du%7D%7B%5Cln%20%5Cleft%285%5Cright%29%7D%5C%3A%5C%5C%5C%5C%5Cmathrm%7Bin%5C%3ALambert%5C%3Aform%7D%3A%5Cquad%20%5Cfrac%7Be%5E%7B%5Cfrac%7B2u%2B3%5Cln%20%5Cleft%285%5Cright%29%7D%7B2%7D%7Du%7D%7Be%5E%7B%5Cfrac%7B3%5Cln%20%5Cleft%285%5Cright%29%7D%7B2%7D%7D%7D%3D%5Cfrac%7B%5Cln%20%5Cleft%285%5Cright%29%7D%7B2e%5E%7B%5Cfrac%7B3%5Cln%20%5Cleft%285%5Cright%29%7D%7B2%7D%7D%7D)
![\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x](https://tex.z-dn.net/?f=%5Cmathrm%7BSolve%5C%3A%7D%5C%3A%5Cfrac%7Be%5E%7B%5Cfrac%7B2u%2B3%5Cln%20%5Cleft%285%5Cright%29%7D%7B2%7D%7Du%7D%7Be%5E%7B%5Cfrac%7B3%5Cln%20%5Cleft%285%5Cright%29%7D%7B2%7D%7D%7D%3D%5Cfrac%7B%5Cln%20%5Cleft%285%5Cright%29%7D%7B2e%5E%7B%5Cfrac%7B3%5Cln%20%5Cleft%285%5Cright%29%7D%7B2%7D%7D%7D%3A%5Cquad%20u%3D%5Ctext%7BW%7D_0%5Cleft%28%5Cfrac%7B%5Cln%20%5Cleft%285%5Cright%29%7D%7B2e%5E%7B%5Cfrac%7B3%5Cln%20%5Cleft%285%5Cright%29%7D%7B2%7D%7D%7D%5Cright%29%5C%5C%5C%5C%5Cmathrm%7BSubstitute%5C%3Aback%7D%5C%3Au%3D%5Cleft%28x-%5Cfrac%7B3%7D%7B2%7D%5Cright%29%5Cln%20%5Cleft%285%5Cright%29%2C%5C%3A%5Cmathrm%7Bsolve%5C%3Afor%7D%5C%3Ax)
![\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}](https://tex.z-dn.net/?f=%5Cmathrm%7BSolve%5C%3A%7D%5C%3A%5Cleft%28x-%5Cfrac%7B3%7D%7B2%7D%5Cright%29%5Cln%20%5Cleft%285%5Cright%29%3D%5Ctext%7BW%7D_0%5Cleft%28%5Cfrac%7B%5Cln%20%5Cleft%285%5Cright%29%7D%7B2e%5E%7B%5Cfrac%7B3%5Cln%20%5Cleft%285%5Cright%29%7D%7B2%7D%7D%7D%5Cright%29%3A%5C%5C%5Cquad%20x%3D%5Cfrac%7B%5Ctext%7BW%7D_0%5Cleft%28%5Cfrac%7B%5Cln%20%5Cleft%285%5Cright%29%7D%7B2e%5E%7B%5Cfrac%7B3%5Cln%20%5Cleft%285%5Cright%29%7D%7B2%7D%7D%7D%5Cright%29%7D%7B%5Cln%20%5Cleft%285%5Cright%29%7D%2B%5Cfrac%7B3%7D%7B2%7D)
I know that’s a should be one of them
Answer:
The Company AB must sell 116 computers to break even
Step-by-step explanation:
we have
-----> equation A
----> equation B
where
C is the cost function
R is the revenue function
x is the number of computers sold
we know that
Break even is when the cost is equal to the revenue
so
equate equation A and equation B
![1.5x=145+\frac{1}{4}x](https://tex.z-dn.net/?f=1.5x%3D145%2B%5Cfrac%7B1%7D%7B4%7Dx)
Solve for x
subtract 1/4x both sides
![1.5x-\frac{1}{4}x=145](https://tex.z-dn.net/?f=1.5x-%5Cfrac%7B1%7D%7B4%7Dx%3D145)
![1.5x-0.25x=145](https://tex.z-dn.net/?f=1.5x-0.25x%3D145)
![1.25x=145](https://tex.z-dn.net/?f=1.25x%3D145)
Divide by 1.25 both sides
![x=116](https://tex.z-dn.net/?f=x%3D116)
therefore
The Company AB must sell 116 computers to break even
Answer:
$45.50
Step-by-step explanation:
Interest = Principal x rate x time (in years)
I = 1400 x .0325 x 1