Equations don't have minimum or maximum, functions do.
Function y=2n^2+5n-25 has minimum -28.125, has no maximum.
Answer:
x = 10°
Step-by-step explanation:
a). Since, opposite angles of a cyclic quadrilateral are supplementary angles"
Therefore, in cyclic quadrilateral ABDE,
m∠ABD + m∠AED = 180°
110° + m∠AED = 180°
m∠AED = 180° - 110°
= 70°
b). AD = ED [Given]
m∠EAD = m∠AED [Since, opposite angles of equal sides are equal in measure]
m∠EAD = m∠AED = 70°
By triangle sum theorem in ΔABD,
m∠BAD + m∠ABD + m∠ADB = 180°
m∠BAD + 110° + 40° = 180°
m∠BAD = 180 - 150
= 30°
m∠AEB = m∠AED + m∠DAB [By angles addition postulate]
m∠AEB = 70° + 30°
= 100°
By triangle sum theorem in the large triangle,
x° + m∠AEB + m∠EAB = 180°
x° + 100° + 70° = 180°
x = 180 - 170
x = 10°
Let's make things easier by simplifying things.
y = 8 and x = 3 is more likely to be understood as a ratio. So for the rest of the answer, their relationship would be represented as y:x
Thus: y:x = 8:3
The problem would be finding y when x = 45
Let us proceed on using the previous equation and substitute x with 45 which would look like this:
y:45 = 8:3
Ratios can also be expressed as fractions which would make things more understandable and easy to solve. So the new form of our equation would be like this:
y/45 = 8/3
Then we proceed with a cross multiplication where the equation becomes like as what is shown below:
3y = 45 * 8
From there, you can solve it by multiplying 45 and 8 then dividing the product with 3 to get y
3y = 360
y = 120
Another way of looking at the problem, especially problems like these, is to take the whole question or statement as an equation. it would probably look like this:
y = 8 when x = 3 : y = ? when x = 45
This would make you understand what approach you can use to solve the given problem.
I: y=(1/2)x+5
II: y=(-3/2)x-7
substitution:
fancy word for insert the definition of one variable in one equation into the other
-> isolate a variable, luckily y is isolated (even in both equations) already
-> substitute y of II into I (=copy right side of II and replace y in I with it):
(-3/2)x-7=(1/2)x+5
-3x-14=x+10
-3x-24=x
-24=4x
-6=x
-> insert x back into I (or II):
y=(1/2)x+5
=(1/2)*(-6)+5
=-3+5=2
elimination: subtract one equation from the other to eliminate a variable, again y is already isolated->no extra work required
I-II:
y-y=(1/2)x+5-[(-3/2)x-7]
0=(1/2)x+5+(3/2)x+7
0=(4/2)x+12
-12=2x
-6=x
-> insert x back into I (or II):
y=(1/2)x+5
=(1/2)*(-6)+5
=-3+5=2