Answer:
D
Step-by-step explanation:
It must first translate 1 unit to the left because quadrilateral F'G'H'I is one unit away from to y axis. It can then translate across the y-axis.
Hello :
<span>note :
an equation of the
circle Center at the w(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : </span><span>x²+y²-16x+6y+53=0
(</span>x²-16x) +( y²+6y ) +53 = 0
(x² -2(8)x +8² - 8²) +(y² +2(3)x -3²+3² ) +53=0
(x² -2(8)x +8²) - 8² +(y² +2(3)x +3²)-3² +53=0
(x-8)² +( y+3)² = 20
the center is : w(8,-3) and ridus : r = <span>√20</span>
The gradient of the function is constant s the independent variable (x) varies The graph passes through the origin. That is to say when x = 0, y = 0. Clearly A and D pass through the origin, and the gradient is constant because they are linear functions, so they are direct variations. The graph of 1/x does not have a constant gradient, so any stretch of this graph (to y = k/x for some constant k) will similarly not be direct variation. Indeed there is a special name for this function, inverse proportion/variation. It appears both B and C are inverse proportion, however if I interpret B as y = (2/5)x instead, it is actually linear. I believe the answer is C. Hope I helped!
Using the points where the blue line crosses the X and Y axis:
(0,5) ans (-1, 0)
The slope is the change in Y over the change in X:
0-5 / -1 - 0 = -5/-1 = 5
Slope = 5
Y intercept = y1 = 5
Equation = y = 5x+5
<span>4x(x) + 3(2x) and 4x2 + 6x => 4x^2 + 6x and 4x^2 + 6x
5(3x) - 4(3x) and 3x => 15x-12x=3x and 3x
4(3a) - 2(4a) and 4a2 => 12a-8a=4a and 4a^2
3(3a) + a(3a) and 3a2 + 9a => 9a+3a^2 and 9a + 3a^2
Can you spot the one which is different?
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