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2 years ago

7

What is the common difference between successive terms in the sequence? 0.36, 0.26, 0.16, 0.06, –0.04, –0.14, ... –0.1 –0.01 0.0

1 0.1

2 answers:

GaryK [48]2 years ago

5 0

Answer:

-0.1

Step-by-step explanation:

just did it

mart [117]2 years ago

4 0

Answer:

The answer is A -0.1

Step-by-step explanation:

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Joel made some muffins. He gave 1/4 of of the muffins to a neighbor. He took 3/8 of the muffins to school. What fraction of the

dolphi86 [110]

Okay! So 1/4 = 2/8 and 3/8 = 3/8 3/8+2/8= 5/8
3/8 is your answer

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What is the point-slope form of a line with a slope -3 that contains the point (10,-1)

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<h2>y + 1 = -3(x - 10)</h2>

Here's the formula:

y - y1 = m(x - x1)

Substitute numbers accordingly:

y1: so 1 goes in the y1 spot (you switch the signs because it was already negative)

x1: and 10 goes to the x1 spot

m: -3 belongs in m

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2 years ago

Why is -7 an integer but not a natural number?

MrMuchimi

Answer:

Natural Numbers are basically "counting numbers". If I ask you to start counting, you'll start at 1, then 2, then 3, and so on. It doesn't include 0 or any negative numbers because you wouldn't start from -10 when asked to count.

<h3>brainliest is appreciated :)</h3>

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3 years ago

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2 years ago

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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