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3 years ago

Find the area of trapezoid

Mathematics

1 answer:

MAXImum [283]3 years ago

8 0

Answer:

1725ft^2

Step-by-step explanation:

I have made it in above picture

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Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8

Kitty [74]

Substitute $z=\ln x$ , so that

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}$

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)$

Then the ODE becomes

$x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0$
$\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0$

which has the characteristic equation $r^2+1=0$ with roots at $r=\pm i$ . This means the characteristic solution for $y(z)$ is

$y_C(z)=C_1\cos z+C_2\sin z$

and in terms of $y(x)$ , this is

$y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)$

From the given initial conditions, we find

$y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1$
$y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8$

so the particular solution to the IVP is

$y(x)=\cos(\ln x)+8\sin(\ln x)$

4 0

3 years ago

Order -6 1/4, -6.35, -6 1/5, and 6.1 from greatest to least. Explain.

san4es73 [151]

If you turn all the numbers into decimals it becomes, -6.25, -6.35, -6.20, and, 6.1 so in order it would go 6.1, -6.20, -6.25, -6.35

3 0

3 years ago

What is the area of rhombus ABCD? Enter your answer in the box. Do not round at any steps. put the answer in units²

AlladinOne [14]

Answer:

Therefore the Area of Rhombus ABCD is 36 unit².

Step-by-step explanation:

Given:

ABCD is a Rhombus

A = (-1,0)

B = (5,-3)

C = (-1,-6)

D = (-7 ,-3)

To Find:

Area of Rhombus ABCD = ?

Solution:

We know that Area of Rhombus is given as

$\textrm{Area of Rhombus}=\dfrac{1}{2}\times d_{1}\times d_{2}$

Where ,

d₁ and d₂ are the Diagonals.

We have,

Diagonals as AC and BD,

Using Distance Formula we get

$l(AC) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}$

Substituting coordinates A and C we get

$l(AC) = \sqrt{((-1-(-1))^{2}+(-6-0)^{2} )}=\sqrt{36}\\l(AC)=6\ unit$

Similarly for BD we have,

$l(BD) = \sqrt{((5-(-7))^{2}+(-3-(-3))^{2} )}=\sqrt{144}\\l(BD)=12\ unit$

Now Substituting AC and BD in Formula we get

$\textrm{Area of Rhombus}=\dfrac{1}{2}\times AC\times BD$

$\textrm{Area of Rhombus}=\dfrac{1}{2}\times 6\times 12=36\ unit^{2}$

Therefore the Area of Rhombus ABCD is 36 unit².

6 0

3 years ago

Question 2 A B C D E F G H

sleet_krkn [62]

Step-by-step explanation:

a. 2(y-8)

= 2y-16

b. 3(x-5)

= 3x-15

c.6(b-4)

= 6b-32

d.7(d-2)

= 7d-14

e.5(2-y)

= 10-5y

f.3(4-t)

= 12-3t

g.5(b-a)

=5b-5a

h.7(2-h)

= 14-7h

4 0

3 years ago

The marketing director of a large department store wants to estimate the average number of customers who enter the store every f

Dahasolnce [82]

Answer:

$36.5674\leq x'\leq61.4326$

Step-by-step explanation:

If we assume that the number of arrivals is normally distributed and we don't know the population standard deviation, we can calculated a 95% confidence interval to estimate the mean value as:

$x-t_{\alpha /2}\frac{s}{\sqrt{n} } \leq x'\leq x-t_{\alpha /2}\frac{s}{\sqrt{n} }$

where x' is the population mean value, x is the sample mean value, s is the sample standard deviation, n is the size of the sample, $\alpha$ is equal to 0.05 (it is calculated as: 1 - 0.95) and $t_{\alpha /2}$ is the t value with n-1 degrees of freedom that let a probability of $\alpha/2$ on the right tail.

So, replacing the mean of the sample by 49, the standard deviation of the sample by 17.38, n by 10 and $t_{\alpha /2}$ by 2.2621 we get:

$49-2.2621\frac{17.38}{\sqrt{10} } \leq x'\leq 49+2.2621\frac{17.38}{\sqrt{10} }\\49-12.4326\leq x'\leq 49+12.4326\\36.5674\leq x'\leq61.4326$

Finally, the interval values that she get is:

$36.5674\leq x'\leq61.4326$

8 0

3 years ago

Find the area of trapezoid​

Find the area of trapezoid