Answer:
14
Step-by-step explanation:
We solve this by using 2 to open the bracket
(3+4)2 is the same as 2(3+4)
2(3)+2(4)
=6+8
=14
OR
(3+4)2 is the same as 2(3+4)
So we can solve the one inside the bracket first
=2(7)=14
Answer:
y = -¼│x − 5│+ 3
Step-by-step explanation:
y = a│x − h│+ k
(h, k) is the vertex of the absolute value graph. In this case, it's (5, 3).
y = a│x − 5│+ 3
One point on the graph is (1, 2). Plug in to find the value of a.
2 = a│1 − 5│+ 3
2 = 4a + 3
a = -¼
Therefore, the graph is:
y = -¼│x − 5│+ 3
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:
y=3/2x+1
Step-by-step explanation:
y-4= -⅔(x-6)
y-4=-⅔x+4
y=-⅔x+4+4
(equation of line 1) y= -⅔x+8 gradient= -⅔
(line 2)gradient=3/2
note* the gradients of perpendicular lines multiplied result to -1
gradient=<u>y²-y²</u>
<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>x²-x¹
<u>3</u><u> </u>=<u>y</u><u>+</u><u>2</u>
<u> </u><u> </u>2. x+2
multiply both sides by 2(x+2)to remove the denominators
3(x+2)=2(y+2)
3x+6=2y+4
3x+6-4=2y
3x+2=2y
divide all sides by 2
3/2x+1=y
y=3/2x+1
Answer:
0.75
Step-by-step explanation:
4.50 is 6% of 75% then 0.75 is 75% in decimal form, so 0.75 is 6% of 4.50
Hope I helped :)
