Solve for x.

2 answers:

7 0

$-\dfrac{2}{3}(3x-4)+3x=\dfrac{5}{6}\ \ \ \ |\text{use distributive property}\\\\-\dfrac{2}{\not3_1}\cdot\not3^1x+\dfrac{8}{3}+3x=\dfrac{5}{6}\\\\-2x+3x+\dfrac{8}{3}=\dfrac{5}{6}\ \ \ \ |-\dfrac{8}{3}\\\\x=\dfrac{5}{6}-\dfrac{8}{3}\\\\x=\dfrac{5}{6}-\dfrac{8\cdot2}{3\cdot2}\\\\x=\dfrac{5}{6}-\dfrac{16}{6}\\\\\boxed{x=-\dfrac{11}{6}}$

Other method:

$-\dfrac{2}{3}(3x-4)+3x=\dfrac{5}{6}\ \ \ \ |\cdot6\\\\2(-2)(3x-4)+18x=5\\\\-4(3x-4)+18x=5\ \ \ \ |\text{use distributive property}\\\\-12x+16+18x=5\ \ \ \ |-16\\\\6x=-11\ \ \ \ |:6\\\\\boxed{x=-\dfrac{11}{6}}$

Vikentia [17]2 years ago

5 0

Answer:

x= -11/6

Step-by-step explanation:

I took the test!

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Click here please. The problems are in picture format

padilas [110]

First picture:

You got the first two right (good job!), so I'll just tackle the third one. If you have to compute $f(x-4)$ , it means that you have to substitute $x-4$ in place of $x$ . Differently from the first two points, this will generate a new function, rather than a specific value:

$f(x) = \sqrt{x+4}+2 \implies f(x-4) = \sqrt{(x-4)+4}+2 = \sqrt{x}+2$

Second picture:

Given the point $(x,y)$ highlighted in the picture, you can deduce that the base is $2x$ units long (since it spans from $(-x,0)$ to $(x,0)$ ) and the height is $y$ units long (because it spans from $(x,0)$ to $(x,y)$ ). So, the area of the rectangle is the multiplication between base and height:

$A = 2xy$

But we know that $y=f(x)=\sqrt{36-x^2}$ , so we have

$A = 2x\sqrt{36-x^2}$

The domain of this function is given by the domain of the square root: we want its argument to be non, negative, so we have

$36-x^2 \geq 0 \iff x^2 \leq 36 \iff -6 \leq x \leq 6$

But since the problem is symmetric, the answer is

$0 \leq x \leq 6$

You can only see the answer $0 < x < 6$ because, if you choose $x=0$ or $x=6$ , the rectangle degenerates to a segment, and your exercise doesn't like this scenario, apparently