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3 years ago

13

13 ft :10 ft 18 ft Find the area of the trapezoid. [? ] square feet Hint: The formula for the area of a trapezoid is: (b17b2). h

1 answer:

Liono4ka [1.6K]3 years ago

3 0

The area is 29473
i just need points

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Paula tiene una cartulina de 50x65 cm. Quiere hacer 5 rectángulos de 210x297 mm. ¿Tendrá suficiente con una cartulina?

brilliants [131]

Answer:

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3 years ago

PLEASE HELP ME PLEAESESES

Bogdan [553]

Answer:

Make a coordinate plane and you have to solve around the orgin say 4,4 and 4,1

Step-by-step explanation:

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3 years ago

I need help ASAP!!!!!

ycow [4]

Input Data :

Point 1 ( x A , y A ) = (-6, -2)

Point 2 ( x B , y B ) = (1, -6)

Objective :

Find the distance between two given points on a line.

Formula :

Distance between two points = √ ( x B − x A )2 + ( y B − y A ) 2

Solution :

Distance between two points = √ ( 1 - -6 ) 2 + ( − 6 − − 2 ) 2

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Distance between points (-6, -2) and (1, -6) is 8.0623

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3 years ago

How can you tell if an equation has direct variation? (Direct variation means proportional)

mario62 [17]

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3 years ago

Find dx/dt when y=2 and dy/dt=1, given that x^4=8y^5-240<br><br> dx/dt=

katrin2010 [14]

Answer:

The value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

Step-by-step explanation:

The given equation is

$x^4=8y^5-240$

We need to find the value of $\frac{dx}{dt}$ .

Differentiate with respect to t.

$4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0$ $[\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0]$

$4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}$

It is given that y=2 and dy/dt=1, substitute these values in the above equation.

$4x^3\frac{dx}{dt}=40(2)^4(1)$

$4x^3\frac{dx}{dt}=40(16)(1)$

$4x^3\frac{dx}{dt}=640$

Divide both sides by 4x³.

$\frac{dx}{dt}=\frac{640}{4x^3}$

$\frac{dx}{dt}=\frac{160}{x^3}$

Therefore the value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

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3 years ago