Since both are equivalent to y, the equations must be equivalent.
x^2-x-3= -3x+5
x^2+2x-8=0
(x+4)(x-2)=0
x=-4, x=2
Plug the values of x in to either equation
y=-3(-4)+5
y= 12+5
y=17
y= -3(2)+5
y=-6+5
y=-1
Final answer: (-4,17) and (2,-1)
Answer:
y=x-6
Step-by-step explanation:
The x-intercept is found replacing y = 0, as follows:
<u>Function:</u> y = -3x + 8
x-intercept:
0 = -3x + 8
3x = 8
x = 8/3
<u>Function</u>: y = -3x + 6
x-intercept:
0 = -3x + 6
3x = 6
x = 6/3 = 2
<u>Function</u>: y = -x - 8
x-intercept:
0 = -x - 8
x = -8
<u>Function</u>: y = x - 6
x-intercept:
0 = x - 6
x = 6
The last option is the only one in which x-intercept is x = 6.
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.
Answer:
E. Each trial is independent
Step-by-step explanation:
I'm not completely sure why all binomial distribution trials are independent, but there are requirements for binomial distributions.
These requirements are:
- Each outcome is either a success (p) or a failure (Q)
- All trials are independent
- There are a fixed number of "n" trials
- The probability of success (p) is the same for each trial
I also just took the test and got this right.
Answer:
69
Step-by-step explanation:
so x2 will give you 9 times 7 which is 63. add it to 2 times 3 which is 63 + 6 will give you 69.
