I golf club charges $10 to join and $5 per game: So, the cost of playing golf is equal to $10 (joining price) plus the product of the number of games played (x) and the price per game (5).

That expression must be higher than 25, since john paid more than $25.

Mathematically speaking

10+ 5 x > 25

Solving for x

5x >25-10

5x >15

x >15/5

x> 3

He could have played more than 3 games, since his cost was higher ( not equal or higher) than 25.

4 0

3 years ago

4^4x-5= 8^3x-4 solve

Alexxandr [17]

$4^{4x-5}=8^{3x-4}\\\\(2^2)^{4x-5}=(2^3)^{3x-4}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\2^{2(4x-5)}=2^{3(3x-4)}\iff2(4x-5)=3(3x-4)\qquad\text{use distributive property}\\\\(2)(4x)+(2)(-5)=(3)(3x)+(3)(-4)\\\\8x-10=9x-12\qquad\text{add 10 to both sides}\\\\8x=9x-2\qquad\text{subtract 9x from both sides}\\\\-x=-2\to \boxed{x=2}$

8 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago

Plss I need answer fast!!

fenix001 [56]

Answer:

Step-by-step explanation:

The equation for the slope of a line is y2-y1/x2-x1, and both triangles have the same slope equation for their hypotenuse.

7 0

2 years ago

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