Answer:
(x + 5)(2x + 1)
Step-by-step explanation:
Given
2x² + 11x + 5
Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 2 × 5 = + 10 and sum = + 11
The factors are + 10 and + 1
Use these factors to split the x- term
2x² + 10x + x + 5 ( factor the first/second and third/fourth terms )
= 2x(x + 5) + 1 (x + 5) ← factor out (x + 5 from each term
= (x + 5)(2x + 1) ← in factored form
Answer:
(letters are going in order from top to bottom)
<u>Statistical</u> <u>Nonstatistical</u>
A
B
C
D
E
F
M to millimetres is 3 steps. m > dm > cm > mm. since it is 3 steps down, the comma goes 3 places left. 4 > 0,4 > 0,04 > 0,004. so 4 m is equal to 0,004 mm
Anything that is inside the perenthasis are just like a regular math problem, so your answer should be:
C. 5
Answer:
A 90
Step-by-step explanation:
multiple ways to prove this.
e.g. since the angle between the two lines from the center of the circle to the 2 tangent touching points is 90 degrees (that is the meaning of these 90 degrees here as the angle of the circle segment defined by the 2 tangent touching points and the circle center), the tangents have the same "behavior" as tan and cot = the tangents at the norm circle at 0 and 90 degrees. they hit each other outside of the circle again at 90 degrees.
another way
imagine the two right triangles of the tangents crossing point to the circle center and the tangent/circle touching points.
the Hypotenuse of each triangle is cutting the 90 degree angle of the circle segment exactly in half (due to the symmetry principle). so the angle between radius side and Hypotenuse is 90/2 = 45 degrees.
that means also the angle of such a right triangle at the tangent crossing point is 45 degrees (as the sum of all angles in a triangle must be 180, we have the remainder of 180 - 90 - 45 = 45 degrees).
the angles of both right triangles at that point are the same, and so we can add 45+45 = 90 degrees for the total angle at the tangent crossing point.
