Answer:
73 degrees.
Step-by-step explanation:
So, let's see the information provided in the question.
At 8 AM, the thermometer was reading -5 degrees.
At 12:30 PM, the temperature has increased by 11 degrees, so it's now +6 degrees outside.
Inside Susan's house, the temperature is 7 degrees more than 11 times the outside temperature, at 12:30 PM.
So, we know the outside temperature at 12:30 PM is +6 degrees.
11 times 6 = 66 degrees.
Then we add the additional 7 degrees, to get a total of 73 degrees.
3x + 2y = 39
5x - y = 13
Multiplying the last by two:
10x - 2y = 26
Adding to the first
13x = 65
x = 5
y = 5x-13 =25-13=12
Check:
3(5)+2(12)=39 good
5(5)-12=13 good
Answer: x=5, y=12
Answer:
since there's only 100 students how can 60 participate in sport and 50 participate in music, that must mean there is 110 students right?
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:
a
Step-by-step explanation:
the answer is a. 6 faces 12 edges and 8 vertices
