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ANEK [815]
3 years ago
15

The graph of a non-invertible function passes through the points (- 1, 2), (0, 6) and (3, - 4) How is the inverse of this functi

on graphed ? Drag a phrase or group of coordinates into each box to correctly complete the statement.

Mathematics
1 answer:
KATRIN_1 [288]3 years ago
8 0

Answer:

The order of ordered pairs of a function and its inverse reverse. The graph of the inverse of this function passes through the points (2,-1), (6,0), (-4,3).

Explanation:

I just took the test.

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5x – 2= 8<br> Please help me
Marta_Voda [28]

Answer:

x = 2 because 5 times 2 = 10 and 10 - 2 = 8

Step-by-step explanation:

6 0
3 years ago
Rewrite the following equation in standard form.<br> y + 7 = 2(x - 6)
babunello [35]

Answer:

(-5)

Step-by-step explanation:

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4 0
3 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
Simplify -3p3 + 5p + (-2p 2) + (-4) - 12p + 5 - (-8p3). Select the answer in descending powers of p.
BigorU [14]
-3p^3 + 5p -2p^2 - 4 - 12p + 5 + 8p^3
5p^3 - 2p^2 - 7p + 1

It is the second choice
7 0
4 years ago
Y = -3 x + 9 <br><br> Y= x -9<br><br> Y = 1/3x + 3 <br><br> Y = -3x -9
sashaice [31]

the correct answer is a

8 0
3 years ago
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