What is the area of the trapezoid? Plz help! :)

Help Pls Help Pls Help Pls

aalyn [17]

Answer:

169

Step-by-step explanation:

13^2 is really just 13 times 13 which equals 169

4 0

4 years ago

Read 2 more answers

Find general solutions of the differential equation. Primes denote derivatives with respect toÂ x.

zepelin [54]

Answer:

$\mathbf{3x^2y^3+2xy^4=C}$

Step-by-step explanation:

From the differential equation given:

$6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0$

The equation above can be re-written as:

$6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0$

$(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0$

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

$\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial x}}} \ \ \text{at each point of R}$

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

$M(x,y) = 6xy^3 +2y^4\ and \ N(x,y) = 9x^2 y^2 +8xy^3$

So;

$\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3$

$\dfrac{\partial N}{\partial y }$

Let's Integrate $\dfrac{\partial F}{\partial x}= M(x,y)$ with respect to x

Then;

$F(x,y) = \int (6xy^3 +2y^4) \ dx$

$F(x,y) = 3x^2 y^3 +2xy^4 +g(y)$

Now, we will have to differentiate the above equation with respect to y and set $\dfrac{\partial F}{\partial x}= N(x,y)$ ; we have:

$\dfrac{\partial F}{\partial y} = \dfrac{\partial}{\partial y } (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0 \\ \\ g(y) = C_1$

Hence, $F(x,y) = 3x^2y^3 +2xy^4 +g(y) \\ \\ F(x,y) = 3x^2y^3 + 2xy^4 +C_1$

Finally; the general solution to the equation is:

$\mathbf{3x^2y^3+2xy^4=C}$

5 0

3 years ago

A pyramid has a square base that is 160 m on each side. what is the perimeter in kilometers

11Alexandr11 [23.1K]

1km=1000m 160m=0.160x4=0.64

3 0

3 years ago

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That-was not what the question was asking.

Leona [35]

Ok what was it then...I could answer it....well depends what type of question it is

7 0

3 years ago

Sharon is jumping from an 18-foot diving board with an initial upward velocity of 4 ft/sWhen Sharon jumps, Megan throws a beach

likoan [24]

Answer:

1.08 seconds

Step-by-step explanation:

given data

first equation:

According to the data, sharon position will be

h(t)= -16 $t^{2}$ + 4t +18---->eq(1)

Second equation;

When Sharon jumps, Megan throws a beach ball up to Sharon with an initial upward velocity of 16 ft/s from a height 5 feet off the ground to the nearest hundredth of a second

therefore, ball position would be

h(t)= -16 $t^{2}$ +16t +5 --->eq(2)

In order to find how long after she jumps does the ball reach Sharon i.e time, we will equate eq(1) and eq(2)

Therefore,

eq(1)= eq(2)

-16 $t^{2}$ +4t +18 = -16 $t^{2}$ + 16t +5 --->(further simplifying it)

-16 $t^{2}$ + 16 $t^{2}$ + 18 -5 = 16t - 4t

13 = 12t

t= 13/12

t= 1.08 s

Thus it took 1.08 seconds.

7 0

3 years ago