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2 years ago

Quadrilateral wxyz is a paralelogram if w=50 what is wzy

Mathematics

1 answer:

slega [8]2 years ago

8 0

It means different digits.
W=50 would count as one of the digits.
Now you have to figure out what “XYZ” is.

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real estate julie bought a house for $100,00 five years ago. if the value of the house has appreciated 5% per year , how much is

Stels [109]

To find the answer, turn the appreciated percent into a decimal:

5% ----> .05 (Divide by 100)

Then multiply it by the total value:

100,000 x .05 = 5000

Then add that to the total:

105,000

So the house is worth a total of $105,000.

Hope this helps!

5 0

3 years ago

Make 10 equations for 50,000

Kruka [31]

Answer:

1. 50 * 1,000 = 50,000

2. 49,001 + 999 = 50,000

3. 5 * 10,000 = 50,000

4. 100,000/2 = 50,000

5. 20,000 + 30,000 = 50,000

6. 50,000/1 = 50,000

7. 90,000 - 40,000 = 50,000

8. 10 + 49,990 = 50,000

9. 5,000 * 10 = 50,000

10. 1,000,000/20 = 50,000

4 0

3 years ago

Read 2 more answers

Wetting the cloth cover of your canteen on a hot day is a good or bad idea

Lyrx [107]

Bad idea.. because the cloth will get hot and then make your water inside warm

8 0

3 years ago

Read 2 more answers

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

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

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

3 years ago

Write an inequality to represent the two amounts , Gita used 6.18 pounds of strawberries and 6.1 pounds of raspberries to make j

Ganezh [65]

Answer:

Strawberry > Rasberry

6.18 > 6.1

Step-by-step explanation:

Data provided in the question

Strawberry = 6.18 pound

Raspberry = 6.1 pound

Based on the above information

As we can derive that the number of pounds in strawberry is greater than the number of pounds in Rasberry

So, equality would be represented as a

a. Strawberry > Rasberry

6.18 > 6.1

Therefore we use the greater sign to show the inequality

5 0

3 years ago