Answer: 90%
Step-by-step explanation:
54/60=0.9 0.9*100= 90
<span>y = slope*x + y-intercept;
</span>We can rewrite our equation in a shorter form : y = mx + b;
y = x + 2 ; m1 = 2 and b1 = 2;
y = -x + 6; m2 = -1 and b2 = 6;
<span>Set the two equations for y equal to each other:
</span>x + 2 = -x + 6 ;
<span>Solve for x. This will be the x-coordinate for the point of intersection:
</span>2x = 4;
x = 2;
<span>Use this x-coordinate and plug it into either of the original equations for the lines and solve for y. This will be the y-coordinate of the point of intersection:
</span>y = 2 + 2 ;
y = 4;
<span>The point of intersection for these two lines is (2 , 4).</span>
Answer:
$2,462,825.99 is the interest accrued
Step-by-step explanation:
hope this helps
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.
Graph of f(x-3) is compressed by a factor of 
horizontally of f(x).
<u>Step-by-step explanation:</u>
We have, the graph of f(x)= 
, on replacing f(x) by f(x-3) we get:
= 
.Below shown are the images for graph of f(x) and f(x-3). Both are functions are exponential , and so having exponential graph but f(x-3) is compressed by a factor of horizontally . Domain and range of both functions are same i.e. F(x) & f(x-3) domain & range are same , just difference in graph : 
.
