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77julia77 [94]
3 years ago
11

Hxxjjxjjjjjjjjjjjjjbxxjxjxxx

Mathematics
1 answer:
Taya2010 [7]3 years ago
7 0

Answer:

xxxjxjxxbjjjjjjjjjjjjxjjxxh

Step-by-step explanation:

Listen that question is easy how u aint get it

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Explain the step(s) needed to solve 4+ p = 1 and then solve for p
liberstina [14]

Answer:

p = -3

Step-by-step explanation:

Apply algebra to the equation by subtracting 4 from each side, that way it will cancel out at least the 4 next to p. It should now look like this: p = -3

Because we can't make anymore moves and the expression clearly states what p is, we're done!

4 0
3 years ago
What are the answers for the chart in number 4,5,6
svetoff [14.1K]
#4(1)=-1. (2)=5. (3)=14. (4)=26 srry but all i know hoped i helped

[how i did]

the first one i did 3×-5+14 which equals -1
and the second one is the same two 3×-3+14=5 you just add the the number on B
7 0
2 years ago
What is cos of negative pi/3?<br> what is sin of negative pi/3?
stellarik [79]
Pi/3 is equivalent to 60 degrees, as 2pi is equal to 360 degrees. cos(60) in a triangle yields 1/2, and sin(60) yields (3^(1/2))/2. Thus, -pi/3, or -60 degrees would be a fourth quadrant point on the unit circle and these values would be negative as well, at cos(-pi/3)=-1/2 and sin(-pi/3)=-(3^(1/2))/2 
4 0
3 years ago
Read 2 more answers
If h(x) = 3 + 2f(x) , where f(1) = 3 and f '(1) = 4, find h'(1).
Leviafan [203]
H'(x) = 2f'(x)

h'(1) = 2*f'(1) = 2*4
h'(1) = 8
3 0
3 years ago
Please help me on this problem
Anna71 [15]

Answer:

The pairs of integer having two real solution forax^{2} -6x+c = 0 are

  1. a = -4, c = 5
  2. a = 1, c = 6
  3. a = 2, c = 3
  4. a = 3, c = 3

Step-by-step explanation:

Given

ax^{2} -6x+c = 0

Now we will solve the equation by putting all the 6 pairs so we get the  following

-3x^{2} -6x-5 = 0 for a = -3 , c=-5

-4x^{2} -6x+5 = 0 for a = -4 , c=5

1x^{2} -6x+6 = 0 for a = 1 , c=6

2x^{2} -6x+3 = 0 for a = 2 , c=3

3x^{2} -6x+3 = 0 for a = 3 , c=3

5x^{2} -6x+4 = 0 for a = 5 , c=4

The above  all are Quadratic equations inn general form ax^{2} +bx+c=0

where we have a,b and c constant values

So for a real Solution we must have

Disciminant , b^{2} -4\timesa\timesc \geq 0

for a = -3 , c=-5 we have

Discriminant =-24 which is less than 0 ∴ not a real solution.

for a = -4 , c=5 we have

Discriminant = 116 which is greater than 0 ∴ a real solution.

for a = 1 , c=6 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 2 , c=3 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 3 , c=3 we have

Discriminant =0 which is equal to 0 ∴ a real solution.

for a = 5 , c=4 we have

Discriminant =-44 which is less than 0 ∴ not a real solution.

7 0
3 years ago
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