The values of <span> f(-2), f(-0.5), and f(3) can be solved by substituting the values inside the parentheses to the function to the variable x. We do as follows:
</span><span>1. f(x)=-x+4
</span>f(-2) =-x+4 = 2+4 = 6
f(-0.5) =-x+4 = 0.5 + 4 = 4.5
f(3) =-x+4 = -3+4 = 1<span>
2. f(x)= 3/8x-3
</span>f(-2)= 3/8x-3 = -15/4<span>
f(-0.5) </span>= 3/8x-3 = -51/16<span>
f(3) </span>= 3/8x-3 = -15/8
Hope these answers the question.
Answer:
B
Step-by-step explanation:
In the first part for every x, y is the value of x plus 2, while in the second part for every x. y is 5
Well 10% is 1/3 o 30% so 10% of the money is 1/3 of 30% of the money
27p*1/3=9p
Basically you can graph a function, for example a parabola by following the step pattern 1,35...
If you take the "standard" parabola, y = x², which has it's vertex at the origin (0, 0), then:
<span>➊ one way you can use a "step pattern" is as follows: </span>
<span>Starting from the vertex as "the first point" ... </span>
<span>OVER 1 (right or left) from the vertex point, UP 1² = 1 from the vertex point </span>
<span>OVER 2 (right or left) from the vertex point, UP 2² = 4 from the vertex point </span>
<span>OVER 3 (right or left) from the vertex point, UP 3² = 9 from the vertex point </span>
<span>OVER 4 (right or left) from the vertex point, UP 4² = 16 from the vertex point </span>
<span>and so on ... </span>
<span>where the "UP" numbers are the sequence of "PERFECT SQUARE" numbers ... </span>
<span>but always counting from the VERTEX EACH time. </span>
<span>➋ another way you can use a "step pattern" is just as you were doing: </span>
<span>Starting with the vertex as "the first point" ... </span>
<span>over 1 (right or left) from the LAST point, up 1 from the LAST point </span>
<span>over 1 (right or left) from the LAST point, up 3 from the LAST point </span>
<span>over 1 (right or left) from the LAST point, up 5 from the LAST point </span>
<span>over 1 (right or left) from the LAST point, up 7 from the LAST point </span>
<span>and so on ... </span>
<span>where the "UP" numbers are the sequence of "ODD" numbers ... </span>
<span>but always counting from the LAST point EACH time. </span>
<span>The reason why both Step Patterns Systems work is that set of PERFECT SQUARE numbers has the feature that the difference between consecutive members is the set of ODD numbers. </span>
<span>For your set of points, the vertex (and all the others) are simply "down 3" from the "standard places": </span>
<span>Standard {..., (-3, 9), (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (3, 9), ...} </span>
<span>shift ↓ 3 : {..., (-3, 6), (-2, 1), (-1, -2), (0, -3), (1,-2), (2, 1), (3, 6), ...} </span>