Answer:
x= (y-5)/2
Step-by-step explanation:
Isolate the x variable.
y= 2x +5
Subtract 5 from both sides.
y-5=2x
Divide both sides by 2.
(y-5)/2= x
Hello from MrBillDoesMath!
Answer: A, C, D
Discussion:
The solution of the quadratic equation ax^2 + bx + c = 0 is
x = ( -b +\- (b^2 - 4ac) ^ (1/2) ) /2a
The expression b^2 - 4ac is called the discriminant, occurs in the numerator of the solution, and also occurs inside a square root (called a radicand).
The question, while intelligible, should be stated more clearly.
Regards, MrB
Here is what you tally chart/table would look like:
Weight Tallies
1/4 lll
2/4 ll
3/4 lll
4/4 l
Here is what your line plot would look like:
X X
X X X
<u>X X X X
</u><u />1/4 2/4 3/4 4/4
Answer:
Area of trapezium = 4.4132 R²
Step-by-step explanation:
Given, MNPK is a trapezoid
MN = PK and ∠NMK = 65°
OT = R.
⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).
Now, sum of interior angles in a quadrilateral of 4 sides = 360°.
⇒ x + x + 65° + 65° = 360°
⇒ x = 115°.
Here, NS is a tangent to the circle and ∠NSO = 90°
consider triangle NOS;
line joining O and N bisects the angle ∠MNP
⇒ ∠ONS =
= 57.5°
Now, tan(57.5°) = 
⇒ 1.5697 = 
⇒ SN = 0.637 R
⇒ NP = 2×SN = 2× 0.637 R = 1.274 R
Now, draw a line parallel to ST from N to line MK
let the intersection point be Q.
⇒ NQ = 2R
Consider triangle NQM,
tan(∠NMQ) = 
⇒ tan65° =
⇒ QM =
QM = 0.9326 R .
⇒ MT = MQ + QT
= 0.9326 R + 0.637 R (as QT = SN)
⇒ MT = 1.5696 R
⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R
Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).
⇒ A = (
) × (ST)
= (
) × 2 R
= 4.4132 R²
⇒ Area of trapezium = 4.4132 R²