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Dimas [21]
2 years ago
8

Don't know this please help

Mathematics
1 answer:
nikitadnepr [17]2 years ago
6 0
I would say answer number 2
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Solve for x under the assumption that x&lt;0. <br> x-5/x&lt;4
AleksandrR [38]
If you would like to solve the inequality x - 5/x < 4, you can do this using the following steps:

<span>x - 5/x < 4
</span>x^2 - 5 < 4x
x^2 - 4x - 5 < 0
(x - 5) * (x + 1) < 0
x < 0
1. x = 5
2. x = -1

The correct result would be x = -1.
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The graph of F(x) can be stretched vertically and flipped over the x-axis to produce the graph of G(x). If F(x)=x^2, which of th
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To stretch it vertically, you'll need a coefficient greater than 1 in front of the x. If you want to flip the graph over the x axis that coefficient needs to be negative. So it can only be D
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What is the constant of proportionality for the equation y=1 1/2x?​
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Answer:

<h3>hope it helps you see the attachment for further information... </h3>

<h3>regards..... </h3>

<h3>_addy_✨✨</h3>

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musickatia [10]

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8 0
2 years ago
Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt
polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt

=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k

The first integral is trivial since (\tan t)'=\sec^2t.

The second can be done by substituting u=t^2-1:

u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C

The third can be found by integrating by parts:

u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t

\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8

\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C

8 0
3 years ago
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