Find factors of -5 that, when added together, will give 4
x² + 4x - 5
x 5
x -1
(x + 5)(x -1)
Using the FOIL method (First, Outside, Inside, Last), we can get the question again.
x(x) = x²
x(-1) = -x
5(x) = 5x
5(-1) = -5
x² - x + 5x - 5
x² + 4x - 5
(x + 5) (x - 1) is your answers
however, i believe it is only asking for one. Therefore, because (x - 1) is a choice, (x - 1) should be your answer
hope this helps
Answer: 2/6 and 1/3
Step-by-step explanation: Equivalent fractions are fractions that have the same value but have different top and bottom numbers.
We can find 1 equivalent fraction by dividing the numerator and the denominator by 2 and we get 2/6. We can also reduce 2/6 by dividing the numerator and the denominator by 2 and we get 1/3.
So two equivalent fractions for 4/12 would be 2/6 and 1/3.
Answer:
this Is 3 and the second one is also 3
Answer:
x > 7/4
Step-by-step explanation:
3(x + 7) < 7(x + 2)
Expand brackets: 3x + 21 < 7x + 14
Subtract 14 from both sides: 3x +7 < 7x
Subtract 3x from both sides: 7 < 4x
Divide both sides by 4: 7/4 < x
Therefore x > 7/4
Answer:
See below
Step-by-step explanation:
(a) Field lines
A negatively charged particle has an electric field associated with it.
The field lines spread out radially from the centre of the point. They are represented by arrows pointing in the direction that a positive charge would move if it were in the field.
Opposite charges attract, so the field lines point toward the centre of the particle.
For an isolated negative particle, the field lines would look like those in Figure 1 below.
If two negative charges are near each other, as in Figure 2, the field lines still point to the centre of charge.
A positive charge approaching from the left is attracted to both charges, but it moves to the closer particle on the left.
We can make a similar statement about appositive charge approaching from the left.
Thus, there are few field lines in the region between the two particles.
(b) Coulomb's Law
The formula for Coulomb's law is
F = (kq₁q₂)/r²
It shows that the force varies inversely as the square of the distance between the charges.
Thus, the force between the charges decreases rapidly as they move further apart.