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3241004551 [841]

2 years ago

13

Can someone help me with this math problem, I will be offering a large amount of points and whoever answers correctly will be ma

rked brainless once 2 people answer on the question! :)

2 answers:

OleMash [197]2 years ago

5 0

Answer:

I believe that this equation would be irrational

Step-by-step explanation:

Aleks04 [339]2 years ago

3 0

The answer is B. Hope this helps :)

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Find the slope of the line that passes through (9,-3) and (9,-8)

miskamm [114]

Answer:

The line has no slope. I hope this helps

7 0

1 year ago

The graph shows the cost of renting skates. Sadie rents a pair of skates at 10 A.M. How much will she pay if she returns the ska

stiv31 [10]

Answer: 1 PM = $8

1:15 PM = $9.0

Step-by-step explanation:

a. Since Sadie rents a pair of skates at 10 A.M and returned the skates by 1 P.M, this means the hours spent was 3 hours. Therefore, based on the graph, for 3 hours rented, he will pay $8

The amount that she will pay if she returns them at 1:15 P.M. will be equal to the value of 3 hours 15 minutes which will be:

= $8 + $1.0

= $9.00

5 0

2 years ago

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}$

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2 years ago

Patterns are made using straight lines and arcs.

KIM [24]

5 rows were added to pattern B

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2 years ago

Is this table a function? why or why not?

gavmur [86]

No because each time it goes up by (1) and in the last one it goes up (2 times) and then decreases (4 times) on the right.

6 0

3 years ago