Answer:
<em>1 and -4</em>
Step-by-step explanation:
<em>The vertical asymptote of a function is gotten by equating the denominator of such function to zero.</em>
Given
f(x) = 4x+8/x^2+3x-4
The vertical asymptotes is expressed as;
x^2+3x-4 = 0
Factorize
x^2+4x-x-4 = 0
x(x+4) -1 (x+4) = 0
(x-1)(x+4) = 0
x-1 = 0 and x+4 = 0
x = 1 and x = -4
<em>Hence the vertical asymptotes of the function are 1 and -4</em>
Answer:
you're not doing anything wrong
Step-by-step explanation:
In order for cos⁻¹ to be a function, its range must be restricted to [0, π]. The cosine value that is its argument is cos(-4π/3) = -1/2. You have properly identified cos⁻¹(-1/2) to be 2π/3.
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Cos and cos⁻¹ are conceptually inverse functions. Hence, conceptually, cos⁻¹(cos(x)) = x, regardless of the value of x. The expected answer here may be -4π/3.
As we discussed above, that would be incorrect. Cos⁻¹ cannot produce output values in the range [-π, -2π] unless it is specifically defined to do so. That would be an unusual definition of cos⁻¹. Nothing in the problem statement suggests anything other than the usual definition of cos⁻¹ applies.
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This is a good one to discuss with your teacher.
Answer:
The amount invested are $2600 and $11400 respectively
Step-by-step explanation:
<em>Let the first amount be x</em>
Given:
(First Investment)
Principal (P1) = x
Rate (R1) = 4.5%
Time (T) = 1 year
(Second Investment)
Principal (P2) = 4x + 1000
Rate (R2) = 6%
Time (T) = 1 year
Income = $801
Calculating the income from the first investment;
![I_1 = \frac{P_1R_1T}{100}](https://tex.z-dn.net/?f=I_1%20%3D%20%5Cfrac%7BP_1R_1T%7D%7B100%7D)
Substitute values for P1, R1 and T
![I_1 = \frac{x * 4.5 * 1}{100}](https://tex.z-dn.net/?f=I_1%20%3D%20%5Cfrac%7Bx%20%2A%204.5%20%2A%201%7D%7B100%7D)
![I_1 = \frac{4.5x}{100}](https://tex.z-dn.net/?f=I_1%20%3D%20%5Cfrac%7B4.5x%7D%7B100%7D)
Calculating the income from the second investment;
![I_2 = \frac{P_2R_2T}{100}](https://tex.z-dn.net/?f=I_2%20%3D%20%5Cfrac%7BP_2R_2T%7D%7B100%7D)
Substitute values for P2, R2 and T
![I_2 = \frac{(4x + 1000) * 6 * 1}{100}](https://tex.z-dn.net/?f=I_2%20%3D%20%5Cfrac%7B%284x%20%2B%201000%29%20%2A%206%20%2A%201%7D%7B100%7D)
![I_2 = \frac{6(4x + 1000)}{100}](https://tex.z-dn.net/?f=I_2%20%3D%20%5Cfrac%7B6%284x%20%2B%201000%29%7D%7B100%7D)
![I_1 + I_2 = Annual\ Income](https://tex.z-dn.net/?f=I_1%20%2B%20I_2%20%3D%20Annual%5C%20Income)
So:
![\frac{4.5x}{100} + \frac{6(4x + 1000)}{100} = 801](https://tex.z-dn.net/?f=%5Cfrac%7B4.5x%7D%7B100%7D%20%2B%20%5Cfrac%7B6%284x%20%2B%201000%29%7D%7B100%7D%20%3D%20801)
Multiply through by 100
![100 * \frac{4.5x}{100} +100 * \frac{6(4x + 1000)}{100} = 801 * 100](https://tex.z-dn.net/?f=100%20%2A%20%5Cfrac%7B4.5x%7D%7B100%7D%20%2B100%20%2A%20%20%5Cfrac%7B6%284x%20%2B%201000%29%7D%7B100%7D%20%3D%20801%20%2A%20100)
![4.5x +6(4x + 1000) = 801 * 100](https://tex.z-dn.net/?f=4.5x%20%2B6%284x%20%2B%201000%29%20%3D%20801%20%2A%20100)
![4.5x +24x + 6000 = 80100](https://tex.z-dn.net/?f=4.5x%20%2B24x%20%2B%206000%20%3D%2080100)
Collect Like Terms
![4.5x +24x = 80100 - 6000](https://tex.z-dn.net/?f=4.5x%20%2B24x%20%3D%2080100%20-%206000)
![28.5x = 74100](https://tex.z-dn.net/?f=28.5x%20%3D%2074100)
Divide through by 28.5
![x = \frac{74100}{28.5}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B74100%7D%7B28.5%7D)
![x = \$2600](https://tex.z-dn.net/?f=x%20%3D%20%5C%242600)
Recall that; the second invest
Amount Invested = 4x + 1000
This gives
![Amount = 4 * \$2600 + 1000](https://tex.z-dn.net/?f=Amount%20%3D%204%20%2A%20%5C%242600%20%2B%201000)
![Amount = \$11400](https://tex.z-dn.net/?f=Amount%20%3D%20%5C%2411400)
Hence;
<em>The amount invested are $2600 and $11400 respectively</em>
Answer:
9 just find the Greatest common multiple
Answer:
{1}
Step-by-step explanation:
We find determinant for the given matrix and set it equal to -1
To find determinant we apply formula
![|A| = a_{11}(a_{22}a_{33} − a_{32}a_{23}) - a_{12}(a_{21}a_{33} − a_{31}a_{23})+a_{13}(a_{21}a_{32} - a_{31}a_{22})](https://tex.z-dn.net/?f=%7CA%7C%20%3D%20a_%7B11%7D%28a_%7B22%7Da_%7B33%7D%20%E2%88%92%20a_%7B32%7Da_%7B23%7D%29%20-%20a_%7B12%7D%28a_%7B21%7Da_%7B33%7D%20%E2%88%92%20a_%7B31%7Da_%7B23%7D%29%2Ba_%7B13%7D%28a_%7B21%7Da_%7B32%7D%20-%20a_%7B31%7Da_%7B22%7D%29)
![|A| = x(1x - 2) - 0(7-7) + 0(14-7x)](https://tex.z-dn.net/?f=%7CA%7C%20%3D%20x%281x%20-%202%29%20-%200%287-7%29%20%2B%200%2814-7x%29%20)
|A| = x^2 - 2x
Now we set the determinant = -1
![x^2 - 2x=-1](https://tex.z-dn.net/?f=x%5E2%20-%202x%3D-1)
Add 1 on both sides
![x^2 - 2x + 1=0](https://tex.z-dn.net/?f=x%5E2%20-%202x%20%2B%201%3D0)
Now factor it
(x-1)(x-1) = 0
Set each factor =0 and solve for x
x- 1=0 so x=1