6 + 3y = 4y +2
3y - 4y = 2 - 6
-y = -4
y = 4
hope that helps :)
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<h3><u>solution</u><u>:</u></h3>
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2/4 esa es una ojalá y pude contestar a tiempo
The coordinates of the pre-image of point F' is (-2, 4)
<h3>How to determine the coordinates of the pre-image of point F'?</h3>
On the given graph, the location of point F' is given as:
F' = (4, -2)
The rule of reflection is given as
Reflection across line y = x
Mathematically, this is represented as
(x, y) = (y, x)
So, we have
F = (-2, 4)
Hence, the coordinates of the pre-image of point F' is (-2, 4)
Read more about transformation at:
brainly.com/question/4289712
#SPJ1
Answer:
To solve the first inequality, you need to subtract 6 from both sides of the inequality, to obtain 4n≤12. This can then be cancelled down to n≤3 by dividing both sides by 4. To solve the second inequality, we first need to eliminate the fraction by multiplying both sides of the inequality by the denominator, obtaining 5n>n^2+4. Since this inequality involves a quadratic expression, we need to convert it into the form of an^2+bn+c<0 before attempting to solve it. In this case, we subtract 5n from both sides of the inequality to obtain n^2-5n+4<0. The next step is to factorise this inequality. To factorise we must find two numbers that can be added to obtain -5 and that can be multiplied to obtain 4. Quick mental mathematics will tell you that these two numbers are -4 and -1 (for inequalities that are more difficult to factorise mentally, you can just use the quadratic equation that can be found in your data booklet) so we can write the inequality as (n-4)(n-1)<0. For inequalities where the co-efficient of n^2 is positive and the the inequality is <0, the range of n must be between the two values of n whereby the factorised expresion equals zero, which are n=1 and n=4. Therefore, the solution is 1<n<4 and we can check this by substituting in n=3, which satisfies the inequality since (3-4)(3-1)=-2<0. Since n is an integer, the expressions n≤3 and n<4 are the same. Therefore, we can write the final answer as either 1<n<4, or n>1 and n≤3.
