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3 years ago

I’ll give you brainlest if you solve this!! Pleasee

Mathematics

1 answer:

Mnenie [13.5K]3 years ago

3 0

Answer:

1, -0.5, -1/4, 0, 0.75

Step-by-step explanation:

i have no clue on the second onde

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A certain college graduate borrows $5900 to buy a car. The lender charges interest at an annual rate of 10%. Assuming that inter

Katena32 [7]

This is the answer to your question

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2 years ago

-10.0-(-3.4)<br><br> also you have to simplify your answer

uysha [10]

-13.4 is the answer. And this is simplified form

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3 years ago

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Find the slope of a line traveling through the points (8, -3) and (2, 6)(8,−3)and(2,6).

Sauron [17]

Answer:

not sure if this is what youre looking for but here

y=-3/2x+9

Step-by-step explanation:

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3 years ago

Pls, look at the picture for the question.

finlep [7]

B. 9*10^16 (because in question give it)

B. 3*10^9 (because answer to first question is divided by seconds in one year (365*24*60*60).

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2 years ago

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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