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4 years ago

Find a rational number halfway in between two numbers 4/7 and 1/8

Mathematics

2 answers:

user100 [1]4 years ago

8 0

I think it’s 5/14 not positive but pretty sure

Makovka662 [10]4 years ago

3 0

Answer:

you did good how old are you is smart

Step-by-step explanation:

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EXREAMLY URGENT!! WILL FOREVER THANK YOU!!!! PLS JUST TAKE A LOOK!!!!! 1. What is the ratio of the sides of triangle XYZ?

sdas [7]

Answer:

Dear Laura Ramirez

Answer to your query is provided below

The ratio of triangle XYZ is 1:√3 :2.

Step-by-step explanation:

A 30-60-90 right triangle (literally pronounced "thirty sixty ninety") is a special type of right triangle where the three angles measure 30 degrees, 60 degrees, and 90 degrees. The triangle is significant because the sides exist in an easy-to-remember ratio: 1:√3 :2.

7 0

3 years ago

A shop sells pairs of trainers for £52 each. They make a percentage profit of 30% on each pair. What was the cost of each pair o

DochEvi [55]

Answer:

£67.60

Step-by-step explanation:

30% of 52 is 15.6

52 + 15.6 = 67.6

I hope this helps, please mark Brainliest, thank you!

5 0

3 years ago

What is the length of BD , given that figure ABCD is a rectangle and AC = 26?

VMariaS [17]

What is the length of BD given that figures ABCD is a rectangle and AC=26

6 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

Guys please help i really don't know the answer!!!!!

iVinArrow [24]

Answer:

The answer is Answer choice A

Step-by-step explanation:

Because 4-1=3

3 0

3 years ago

Read 2 more answers