Answer:
Step-by-step explanation:
Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?
Let p(x)=kpx+dp and q(x)=kqx+dq than
f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7
p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7
(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)
So you want:
−2kpk2q=0
and
kpkq=−1
and
2kpd2p−3kpdq+7=0
Now I amfraid this doesn’t work as −2kpk2q=0 that either kp or kq is zero but than their product can’t be anything but 0 not −1 .
Answer: there are no such linear functions.
Tanα=5/12
α=arctan(5/12)°
α≈22.62° (to nearest hundredth of a degree)
<em>The</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>x^</em><em>2</em><em>-</em><em>1</em><em>.</em>
<em>EXPLANATION</em><em>:</em>
<em>To</em><em> </em><em>be</em><em> </em><em>a</em><em> </em><em>polynomial</em><em>,</em><em> </em><em>the</em><em> </em><em>power</em><em> </em><em>of</em><em> </em><em>each</em><em> </em><em>term</em><em> </em><em>must</em><em> </em><em>be</em><em> </em><em>a</em><em> </em><em>whole</em><em> </em><em>number</em><em>.</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>be</em><em> </em><em>helpful</em><em> </em><em>to</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
<em>~</em><em>p</em><em>r</em><em>a</em><em>g</em><em>y</em><em>a</em>
Answer:
-5.5
Step-by-step explanation:
well lets make an equation first. 4x-6<-2 . Now we have to find what x equals.
And now we know that x<1
Answer:
you have to put it on the graph
