Let , coordinate of points are P( h,k ).
Also , k = 3h + 1
Distance of P from origin :
Distance of P from ( -3, 4 ) :
Now , these distance are equal :
Solving above equation , we get :
Hence , this is the required solution.
Answer:
1.a=2
2. C x=2 and x=-3
Step-by-step explanation:
The standard form for the quadratic function is
ax^2 +bx+c
so we need to rewrite the function to be in this form
2x^2 -10 = 7x
Subtract 7x from each side
2x^2 -7x-10 = 7x-7x
2x^2 -7x-10 = 0
a =2, b= -7 c=-10
2. The quadratic formula is
-b ± sqrt(b^2 -4ac)
----------------------------
2a
2x^2 + 2x=12
Lest get the equation in proper form
2x^2 + 2x-12 = 12-12
2x^2 +2x-12 =0
a=2 b=2 c=-12
Lets substitute what we know
-2 ± sqrt(2^2 -4(2)(-12))
----------------------------
2(2)
-2 ± sqrt(4+96)
----------------------------
2(2)
-2 ± sqrt(100)
----------------------------
4
-2 ± 10
----------------------------
4
-2 + 10 -2-10
----------- and --------------
4 4
8/4 and -12/4
2 and -3
Answer:
It is the third option. (the one which has a black point at X=1, Y=1)
Step-by-step explanation:
For simplicity what you should do is, you should satisfy the second condition that is, y=0 if x=1.
Here only the third option is satisfying this condition.
Whenever there is is a white circle or a white point it means that the point does not satisfy the condition and the the function is not defined at that point. And the black point means that the circle is defined at that point and satisfies the condition.
Henceforth, the correct answer is the third option.
A. Write a function f(x) to represent the price after the 80% markup.
<span>b. Write a function g(x) to represent the price after the 25% markdown. </span>
<span>c. Use a composition function to find the price of an item after both price adjustments that originally costs the boutique $150. </span>
<span>d. Does the order in which the adjustments are applied make a difference? Explain.
</span>
answers
<span>a) f(x) = 1.8x
b) f(x) = 0.75(1.8x)
c) f(150) = 0.75(1.8(150) = $202.50
d) No, it doesn't matter. The result is the same.
</span>