
Here, in the given figure. Angle y° forms vertical opposite angle pair with Angle 163°, therefore they have equal values.
that means ~

Answer:
D(1, 2) → D'(2, 7)
E(-3, -5) → E'(-10, 0)
F(4, -1) → F'(11, 4)
Step-by-step explanation:
D(1, 2) → D' ________
Translate image (3x - 1, y + 5)
(1, 2), x = 1 and y = 2
3x - 1 = 3(1) - 1 = 3 - 1 = 2
y + 5 = (2) + 5 = 7
E(-3, -5) → E' ________
(-3, -5), x = -3 and y = -5
3x - 1 = 3(-3) - 1 = -9 - 1 = -10
y + 5 = (-5) + 5 = 0
F(4, -1) → F' ________
(4, -1), x = 4 and y = -1
3x - 1 = 3(4) - 1 = 12 - 1 = 11
y + 5 = (-1) + 5 = 4
9514 1404 393
Answer:
4) 6x
5) 2x +3
Step-by-step explanation:
We can work both these problems at once by finding an applicable rule.

where O(h²) is the series of terms involving h² and higher powers. When divided by h, each term has h as a multiplier, so the series sums to zero when h approaches zero. Of course, if n < 2, there are no O(h²) terms in the expansion, so that can be ignored.
This can be referred to as the <em>power rule</em>.
Note that for the quadratic f(x) = ax^2 +bx +c, the limit of the sum is the sum of the limits, so this applies to the terms individually:
lim[h→0](f(x+h)-f(x))/h = 2ax +b
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4. The gradient of 3x^2 is 3(2)x^(2-1) = 6x.
5. The gradient of x^2 +3x +1 is 2x +3.
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If you need to "show work" for these problems individually, use the appropriate values for 'a' and 'n' in the above derivation of the power rule.