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2 years ago

Write each phrase as an algebraic expression Two times n squared added to nine

Mathematics

2 answers:

Bad White [126]2 years ago

8 0

The answer for this question is 2n^2+9

expeople1 [14]2 years ago

7 0

Answer:

2n^2+9 is the phrase for correct expression

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I need help with this question?????

tangare [24]

Its just multiplied by 2

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3 years ago

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Jackie names the angle < BAC and Lucy names the angle correct? Mr. Joseph draws the angle below on the board: (in the picture

stich3 [128]

Answer:

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Step-by-step explanation:

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3 years ago

Ally is making a scale diagram of her classroom. She uses a scale factor of 3 centimeters per foot to draw the diagram. The actu

Soloha48 [4]

18*3 = 54

20*3 = 60

so the scale drawing would be 54cm x 60cm

6 0

2 years ago

The number can be divided by the least common multiple of 3 and 5

kondaur [170]

9514 1404 393

Answer:

255

Step-by-step explanation:

If the sum of digits is 12, the number is divisible by 3. If the number ends in 0 or 5, it is divisible by 5. So, we're looking for ...

2x0 . . . where x is a digit and 2+x+0 = 12 . . . . . . not possible

2x5 . . . where x is a digit and 2+x+5 = 12 . . . . true for x=5

The number is 255.

3 0

3 years ago

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