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2 years ago

What is the area? 3 in 2 in 4 in

Mathematics

1 answer:

Viktor [21]2 years ago

7 0

Answer:

Step-by-step explanation:

because to find the area you multiply them all together, 3x2x4

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What is the area of a triangle with a base of 4.24 and hight of 3?

Alina [70]

Answer:

6.36

Step-by-step explanation

7 0

3 years ago

4(x+8)-2(x-7)=46 x=?

Scorpion4ik [409]

Answer:

x = 0/2

x = 0

Step-by-step explanation:

4(x+8)-2(x-7) = 46

Expanding the brackets

4x +32 - 2x +14 = 46

2x + 46 = 46

2x = 46-46

2x = 0

x = 0/2

x = 0

5 0

2 years ago

A car travels at a constant speed for 2 hours. what is that speed miles per hour

earnstyle [38]

Answer:

Distance/2 hours

Step-by-step explanation:

Distance=Speed times Time, and since the time is 2 hours, we can say Distance=Speed times 2 hours.

Now we divide both sides by 2 hours, resulting in Distance/2 Hours= Speed.

DUBS

Please consider a brainliest!

3 0

2 years ago

MATHEMATICS: can anybody help or show me how to do this. look at the picture i need help with question 3?

Alekssandra [29.7K]

It wants you to come up with three events. You'll place them on the line right above, where you think they would go.

For example, if your first event was "seeing a unicorn" you would plot it by "impossible." Some things have a fifty fifty chance, or are likely, like you finding a penny. Use any scenario you can think of. How likely is it that it would come true?

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3 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

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

7 0

3 years ago