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3 years ago

Solve the equation 4+6x+5=7x

Mathematics

1 answer:

Leni [432]3 years ago

6 0

The answer to your question is x=9

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Evaluate lim x approaches to 2 : (sqrt(6-x)-2)/(sqrt(3-x)-1)

solong [7]

Answer:

The answer is "0.5".

Step-by-step explanation:

Given:

$\to \lim_{x\to 2} \frac{(\sqrt{(6-x)}-2)}{(\sqrt{(3-x)}-1)}\\\\ \to \lim_{x\to 2} \frac{\frac{d(\sqrt{(6-x)}-2)}{dx}}{\frac{(\sqrt{(3-x)}-1)}{dx}}\\\\$

$\to \lim_{x\to 2} \frac{\frac{-1}{2\sqrt{(6-x)} }}{\frac{-1}{2\sqrt{(3-x)}}}\\\\$

$\to \lim_{x\to 2} \frac{\sqrt{3-x}}{\sqrt{6-x}}\\\\ \to \frac{\sqrt{3-2}}{\sqrt{6-2}}\\\\ \to \frac{\sqrt{1}} {\sqrt{4}}\\\\ \to \frac{1}{2}\\\\ \to 0.5$

5 0

2 years ago

Help Me And Meh Frand Solve This ;-;

pychu [463]

Answer:

1. a (s^7)

2. b (g^11)

Step-by-step explanation:

8 0

3 years ago

What is the answer 4x+3=-5

weqwewe [10]

Answer: x=-2

Step-by-step explanation:

What we are doing is isolating the variable

4x+3=-5

4x-3=-3 subtract 3 from five

4x=-8 divide to solve x

x=-2

5 0

3 years ago

Read 2 more answers

Which of the following is a statistical question?

choli [55]

The answer would be A since in that case its a percentage and there can be multiple answers. In the other questions, there is a definitive answer for it.

Hope this helps :)

5 0

3 years ago

Read 2 more answers

Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x > 0)

vfiekz [6]

Answer:

Step-by-step explanation:

given a point $(x_0,y_0)$ the equation of a line with slope m that passes through the given point is

$y-y_0 = m(x-x_0)$ or equivalently

$y = mx+(y_0-mx_0)$ .

Recall that a line of the form $y=mx+b$ , the y intercept is b and the x intercept is $\frac{-b}{m}$ .

So, in our case, the y intercept is $(y_0-mx_0)$ and the x intercept is $\frac{mx_0-y_0}{m}$ .

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph $(x_0,\frac{1}{x_0})$ . Which means that $y_0=\frac{1}{x_0}$

The slope of the tangent line is given by the derivative of the function evaluated at $x_0$ . Using the properties of derivatives, we get

$y' = \frac{-1}{x^2}$ . So evaluated at $x_0$ we get $m = \frac{-1}{x_0^2}$

Replacing the values in our previous findings we get that the y intercept is

$(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}$

The x intercept is

$\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0$

The triangle in consideration has height $\frac{2}{x_0}$ and base $2x_0$ . So the area is

$\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2$

So regardless of the point we take on the graph, the area of the triangle is always 2.

6 0

3 years ago