It has not been indicated whether the figure in the questions is a triangle or a quadrilateral. Irrespective of the shape, this can be solved. The two possible shapes and angles have been indicated in the attached image.
Now, from the information given we can infer that there is a line BD that cuts angle ABC in two parts: angle ABD and angle DBC
⇒ Angle ABC = Angle ABD + Angle DBC
Also, we know that angle ABC is 1 degree less than 3 times the angle ABD, and that angle DBC is 47 degree
Let angle ABD be x
⇒ Angle ABC = 3x-1
Also, Angle ABC = Angle ABD + Angle DBC
Substituting the values in the above equations
⇒ 3x-1 = x+47
⇒ 2x = 48
⇒ x = 24
So angle ABD = 24 degree, and angle ABC = 3(24)-1 = 71-1 = 71 degree
No, because when adding 1 to any odd number it will become even, and all even numbers besides 2 are not prime. This means that if n was 1 the resulting number would be prime, since 2 is prime.
Answer:
6x5 can also be 6+6+6+6+6
3+3+3+3 can also be 3x4
5+5+5+5+5 can also be 25
3x3 can be 3+3+3
Step-by-step explanation:
#5: This is a function. It passes the vertical line test.
#6: This is not a function. It does not pass the vertical line test. For example, x=0 has six y-values paired with that one x-value. A function that does not make.
#7: This passes the vertical line test.
Answer:
The answer to your question is:
x = 1; y = 1, z = 0
Step-by-step explanation:
-2x + 2y + 3z = 0 (I)
-2x - y + z = -3 (II)
2x + 3y + 3z = 5 (III)
I and II. Multiply II by 2
-2x + 2y + 3z = 0
(2)(-2x - y + z = -3 )
-2x + 2y + 3z = 0
-4x - 2y + 2z = -6
-6x + 5z = - 6 (IV)
II and III. Multiply II by 3
-6x - 3y + 3z = -9 (II)
2x + 3y + 3z = 5 (III)
-4x + 6z = -4 (V)
IV and V. Multiply IV by 2 and V by -3
-12x + 10z = -12
12x - 18z = 12
- 8 z = 0
z = 0
Substitute z in IV
-6x + 5(0) = -6
-6x = -6
x = 1
Substitute x and z in I
-2(1) + 2y + 3(0) = 0
-2 + 2y = 0
2y = 2
y = 1
