Answer:
Option B: 3x + 3y - 3
Step-by-step explanation:
In an isosceles triangle, the two congruent sides are equal.
We are told the base is; x - y - 2 units
Now let each of the congruent sides be represented as A.
Thus the perimeter equation will be;
P = 2A + x - y - 2
Now, we are told that the perimeter is; 7x + 5y - 8 units
Thus;
7x + 5y - 8 = 2A + x - y - 2
Rearranging gives;
7x - x + 5y + y - 8 + 2 = 2A
2A = 6x + 6y - 6
Divide through by 2 to give;
A = 3x + 3y - 3 units
Answer:
Step-by-step explanation:
From the information given:
We are to find; an expression for the total no. of articles written since 1983
∴
The total no. of articles written since 1983 ![=\int \limts ^t_0 U(t) dt](https://tex.z-dn.net/?f=%3D%5Cint%20%5Climts%20%5Et_0%20U%28t%29%20dt)
![= \int ^t_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \ dt](https://tex.z-dn.net/?f=%3D%20%5Cint%20%5Et_0%20%5Climits%20%5Cdfrac%7B4.1%20e%5E%7B0.6%5C%20t%7D%7D%7B0.2%2Be%5E%7B0.6t%7D%7D%20%5C%20%20dt)
![=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^t_0](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B4.1%7D%7B0.6%7D%5Cbigg%20%5B%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%200.2%20%2B%20e%5E%7B0.6%20x%7D%20%5Cbigg%7C%20%5Cbigg%5D%5Et_0)
![=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B4.1%7D%7B0.6%7D%5Cbigg%20%5B%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%200.2%20%2B%20e%5E%7B0.6%20x%7D%20%5Cbigg%7C%20-%20%5Cmathtt%7BIn%7D%20%5Cbigg%7C0.2%2B1%5Cbigg%7C%5Cbigg%5D)
![=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{0.6 x}}{1.2} \bigg|\bigg]](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B4.1%7D%7B0.6%7D%5Cbigg%20%5B%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%20%5Cdfrac%7B0.2%20%2B%20e%5E%7B0.6%20x%7D%7D%7B1.2%7D%20%5Cbigg%7C%5Cbigg%5D)
![=\dfrac{41}{6} \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-\dfrac{41}{6} \mathtt{ In} \bigg | 1.2\bigg |](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B41%7D%7B6%7D%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%20%7B0.2%20%2B%20e%5E%7B0.6%20x%7D%20%5Cbigg%7C-%5Cdfrac%7B41%7D%7B6%7D%20%5Cmathtt%7B%20In%7D%20%5Cbigg%20%7C%201.2%5Cbigg%20%7C)
![=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.2458](https://tex.z-dn.net/?f=%3D6.833%5Ctimes%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%20%7B0.2%20%2B%20e%5E%7B0.6%20x%7D%20%5Cbigg%7C-1.2458)
![=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}](https://tex.z-dn.net/?f=%3D6.833%5Ctimes%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%20%7B0.2%20%2B%20e%5E%7B0.6%20x%7D%20%5Cbigg%7C-1.25%20%5C%20%5Cmathbf%7Bthousand%20%5C%20articles%7D)
Therefore, the total number of articles written since 1983 is ![=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}](https://tex.z-dn.net/?f=%3D6.833%5Ctimes%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%20%7B0.2%20%2B%20e%5E%7B0.6%20x%7D%20%5Cbigg%7C-1.25%20%5C%20%5Cmathbf%7Bthousand%20%5C%20articles%7D)
b. To find how many articles were being written from 1983 to 2003
i.e. t = 2003 - 1983 = 20
∴
Total articles written from 1983 to 2003 is ![=\int \limts ^{20}_0 U(t) dt](https://tex.z-dn.net/?f=%3D%5Cint%20%5Climts%20%5E%7B20%7D_0%20U%28t%29%20dt)
![= \int ^{20}_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \ dt](https://tex.z-dn.net/?f=%3D%20%5Cint%20%5E%7B20%7D_0%20%5Climits%20%5Cdfrac%7B4.1%20e%5E%7B0.6%5C%20t%7D%7D%7B0.2%2Be%5E%7B0.6t%7D%7D%20%5C%20%20dt)
![=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^{20}_0](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B4.1%7D%7B0.6%7D%5Cbigg%20%5B%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%200.2%20%2B%20e%5E%7B0.6%20x%7D%20%5Cbigg%7C%20%5Cbigg%5D%5E%7B20%7D_0)
![=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 *20} \bigg| - \mathtt{In} \bigg|0.2+e^{0.6*0}\bigg|\bigg]](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B4.1%7D%7B0.6%7D%5Cbigg%20%5B%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%200.2%20%2B%20e%5E%7B0.6%20%2A20%7D%20%5Cbigg%7C%20-%20%5Cmathtt%7BIn%7D%20%5Cbigg%7C0.2%2Be%5E%7B0.6%2A0%7D%5Cbigg%7C%5Cbigg%5D)
![=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+e^0\bigg|\bigg]](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B4.1%7D%7B0.6%7D%5Cbigg%20%5B%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%200.2%20%2B%20e%5E%7B12%7D%20%5Cbigg%7C%20-%20%5Cmathtt%7BIn%7D%20%5Cbigg%7C0.2%2Be%5E0%5Cbigg%7C%5Cbigg%5D)
![=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B4.1%7D%7B0.6%7D%5Cbigg%20%5B%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%200.2%20%2B%20e%5E%7B12%7D%20%5Cbigg%7C%20-%20%5Cmathtt%7BIn%7D%20%5Cbigg%7C0.2%2B1%5Cbigg%7C%5Cbigg%5D)
![=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|1.2\bigg|\bigg]](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B4.1%7D%7B0.6%7D%5Cbigg%20%5B%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%200.2%20%2B%20e%5E%7B12%7D%20%5Cbigg%7C%20-%20%5Cmathtt%7BIn%7D%20%5Cbigg%7C1.2%5Cbigg%7C%5Cbigg%5D)
![=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{12}}{1.2} \bigg|\bigg]](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B4.1%7D%7B0.6%7D%5Cbigg%20%5B%20%5Cmathtt%7BIn%7D%20%5Cbigg%20%7C%20%5Cdfrac%7B0.2%20%2B%20e%5E%7B12%7D%7D%7B1.2%7D%20%5Cbigg%7C%5Cbigg%5D)
= 80.75 thousand articles
Your equation is a little messed up comment me the equation you need help with
Answer:
I think the answer is 11. 18
H(k) = k^2 - k so we just substitute in 10 for K.
h(10) = 10^2 - 10. 100 - 10 = 90
h(10) = 90.