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3 years ago

Find a number that is 5 times as much as 4 divided by 2

Mathematics

2 answers:

Marysya12 [62]3 years ago

7 0

Yeah wouldn't it be 10???

barxatty [35]3 years ago

4 0

10....

4 x 5 ÷ 2 is 10

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Answer the photo below thanks

nataly862011 [7]

Answer:

octagon, because 4 + 4 = eight sides, and octagons have 8 sides

Step-by-step explanation:

Hope that helps!

7 0

3 years ago

Read 2 more answers

Find the distance between the point - 3, 2 and 1 - 1

dusya [7]

Answer:

5 units

Step-by-step explanation:

Find the distance between the point - 3, 2 and 1 - 1

Given data

x1= -3

y1= 2

x2= 1

y2= -1

The expression for the distance between two points is

d=√((x_2-x_1)²+(y_2-y_1)²)

subtitute

d=√((1-(-3))²+(-1-(2))²)

d=√((1+3))²+(-1-2))²)

d=√((4)²+(-3))²)

d=√16+9

d=√25

d=5

Hence the distance between the points is 5 units

7 0

2 years ago

Can someone help me do this problem and explain it to me step by step? If so I’ll mark you brainliest

Advocard [28]

Answer:

-21

Step-by-step explanation:

1) if r = -4

Substitute r = -4 in 3r - 9

= 3 * (-4) - 9

= -12 - 9

= -21

2) ( 4 / 7 ) * x = 16

4 * x = 16 * 7

x = 4 * 7

x = 28

7 0

2 years ago

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

2 years ago

I'll mark brainliest if you help me on this

kirill [66]

Answer:

3 3/4 i think

7 0

2 years ago

Read 2 more answers