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3 years ago

HELP PLS !! What is the area of this trapezoid? Enter your answer in the box.

Mathematics

2 answers:

tamaranim1 [39]3 years ago

5 0

Answer:

A =315 cm^2

Step-by-step explanation:

The area of a trapezoid is found by

A = 1/2 (b1+b2) *h

where b1 and b2 are the lengths of the bases and h is the height

A = 1/2 ( 13+29) *15

A = 1/2 (42)* 15

A =315 cm^2

S_A_V [24]3 years ago

4 0

Answer:

$315 {cm}^{2}$

Step-by-step explanation:

$area \\ = \frac{1}{2} \times (a + b) \times h \\ = \frac{1}{2} \times (13 + 29) \times 15 \\ = \frac{1}{2} \times 42 \times 15 \\ = \frac{630}{2} \\ = 315 {cm}^{2}$

hope this helps

brainliest appreciated

good luck! have a nice day!

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Answer:

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Given $f_{XY} (x,y) = c(4x + 2y +1) ; 0 < x < 40\,and\, 0 < y$

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3 years ago

the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i

nekit [7.7K]

Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

$A = \frac{1}{2}\cdot b\cdot h$ (Eq. 1)

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

$\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square centimeters per minute.

$\frac{db}{dt}$ - Rate of change of base in time, measured in centimeters per minute.

$\frac{dh}{dt}$ - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

$\frac{1}{2}\cdot\frac{db}{dt}\cdot h = \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}$

$\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt}$ (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

$b = \frac{2\cdot A}{h}$

If we know that $A = 130\,cm^{2}$ , $h = 15\,cm$ , $\frac{dh}{dt} = 2.5\,\frac{cm}{min}$ and $\frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}$ , the rate of change of the base of the triangle in time is:

$b = \frac{2\cdot (130\,cm^{2})}{15\,cm}$

$b = 17.333\,cm$

$\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)$

$\frac{db}{dt} = -2.262\,\frac{cm}{min}$

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

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