Answer:
standard form for a quadratic expression
Step-by-step explanation:
This is a quadratic expression. If you want this expression in standard form, write the terms in descending order of powers of x: -2x^2 + 1.
This -2x^2 + 1 has not yet been factored.
-2x^2 + 1 is in standard form.
Answer:
3rd option
Step-by-step explanation:
Since the centre of dilatation is at the origin then multiply the coordinates by 
K (- 3, 9 ) → K' (- 3( ), 9( ) ) → K' (- 1, 3 )
L (- 9, 0 ) → L' (- 9( ), 0 ( ) ) → L' (- 3, 0 )
M (2, - 8 ) → M' (2 ( ), - 8 ( ) ) → M' ( 
, - 
)
N (6, 4 ) → N' (6 ( ) , 4 ( ) ) → N' ( 2, 
)
The greatest common factor is 2x^2.
the largest number that divides evenly into <span>10x^5−16x^4+4x^2
is two.
the highest degree is x^2 so the answer would be 2x^2
</span>
Let the regular price be p. Then the new price, after a 35% markdown, is 0.65p.
Mult. $3.29 by 0.65, Result: $2.14. No match here.
Mult. $4.19 by 0.65. Result: $2.72. We've got a match here!
Mult. $2.79 by .65. Result: $1.81. No match here.
Mult. $3.09 by .65. Result: $2.01. Match!
And so on.
Answer:
Ground speed = 131.25 km/h
Bearing = 92.68º
Step-by-step explanation:
If we imagine a triangle to depict the question with one side 134 km/h at an angle of 78º to the right of vertical (the plane's speed vector).
At the end, we have a line 21 km/h at an angle of 180 (from north), a vertical down from the end of the plane's vector. This vertex is angle B
This makes a triangle with the interior angle B of 78º.
The 3rd side is the ground speed which we can find using the Cosine rule:
b² = a² + c² - 2•a•c•cos(78)
b² = 134² + 21² - 2•134•21•cos(78)
b² = 17226.873
b = 131.25 km/h
The true course bearing" is the ground track. Thus;
Using the Sine rule to find angle A, we have;
(sin(A))/34= sin(78)/b
Thus,
sin(A) = (34 x sin(78))/131.25
sin(A) = 0.2534
A = sin^(-1)0.2534
A = 14.68º
Ground track = A + 78 = 14.68 + 78 = 92.68º
