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3 years ago

The functions f(x) and g(x) are shown below f(x) = 5x g(x) = 5x Which statement best describes the graph of f(x) and g(x)?

2 answers:

8 0

The functions supplied appear to be the same? Regardless:

We have the equation y = 5x.
Therefore the gradient of this graph will be 5, so for every 1 increase in the y axis, there will be 5 in the x.

It will appear as a straight line passing through the origin and the point (1, 5).

melamori03 [73]3 years ago

3 0

It is the graphs will both their y-intercepts= to 5

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Solve the equation - 2(3x+5/x+2) -5(x+2/3x-5) = 3 , where x unequal -2 and x unequal 5/3

melomori [17]

Answer:

Step-by-step explanation:

(a - b)(a + b) =a² - b²

(3x + 5)(3x -5) =(3x)² - 5² = 9x² - 25

(x +2)(x+2) = (x+ 2)² = x²+ 2*2*x+2² = x² + 4x + 4

(x+2)(3x -5 ) = x*3x - 5*x +2*3x +2*(-5)

= 3x² -5x +6x -10

= 3x² + x - 10

$\frac{2*(3x+5)}{x+2}-\frac{5(x+2)}{3x-5}=3\\\\\frac{2*(3x+5)*(3x-5)}{(x+2)(3x-5)}-\frac{5*(x+2)(x+2)}{(3x-5)(x+2)}=3\\\\\frac{2*(9x^{2}-25)-5(x^{2}+4x+4)}{(3x-5)(x+2)}=3$

$\frac{18x^{2}-25-5x^{2}-20x-20}{3x^{2}+x-10}=3\\\\\frac{18x^{2}-5x^{2}-20x - 25-20}{3x^{2}+x-10}=3\\\\$

13x² - 20x - 45 = 3*(3x² + x - 10)

13x² - 20x - 45 =9x² + 3x - 30

13x² - 9x² -20x - 3x - 45 + 30 = 0

4x² - 23x - 15 = 0

8 0

3 years ago

Of 14 possible books you plan to take 8 with you on vacation how many possible coalitions of 8 could you have?

Brilliant_brown [7]

Answer:

let me think about this

Step-by-step explanation:

question i may help

7 0

3 years ago

Read 2 more answers

Need help to do this step by step please <br> 3x2 - 2x=-1

sdas [7]

Answer:

Solving the expression $3x^2-2x=-1$ we get: $\mathbf{x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}}$

Step-by-step explanation:

We need to solve the expression: $3x^2-2x=-1$

This is a quadratic expression and it can be solved using quadratic formula

Solving:

$3x^2-2x=-1\\$

we can write it as:

$3x^2-2x+1=0$

The quadratic formula is: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

where a = 3, b = -2 and c= 1

Putting values and solving:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(3)(1)}}{2(3)}\\x=\frac{2\pm\sqrt{4-12}}{2(3)}\\x=\frac{2\pm\sqrt{-8}}{6}\\We\:know\:that\:\sqrt{-1}=i\\x=\frac{2\pm\sqrt{8}\sqrt{-1} }{6} \\We\:know\:\sqrt{8}=\sqrt{2\times 2 \times 2}=\sqrt{2^2 \times 2}=2\sqrt{2} \\x=\frac{2\pm2\sqrt{2}i }{6}\\Now,\\x=\frac{2+2\sqrt{2}i }{6}\:or\:x=\frac{2-2\sqrt{2}i }{6}\\x=\frac{2(1+\sqrt{2}i) }{6}\:or\:x=\frac{2(1-\sqrt{2}i) }{6}\\x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}$