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alexdok [17]
2 years ago
14

What is the coefficient in the expression y + 14?

Mathematics
1 answer:
Lady bird [3.3K]2 years ago
5 0

Answer:

1

Step-by-step explanation:

y + 14

=> 1(y) + 14

=> Coefficient = 1

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What are the coordinates of the vertex of the parabola described by the equation below? y=7(x+5)^2-4
Ipatiy [6.2K]

Answer:

y= 7(x+5)^2 -4

The vertex form for a parabola is given by this expression:

y = a(x-h)^2 +k

By direct comparison we see that for this case:

a = 7, h = -5 and k=-4

And we know from the general expression that the vertex is:

V= (h,k)

So then the vertex for this case is:

V= (-5,-4)

Step-by-step explanation:

For this case we have the following function:

y= 7(x+5)^2 -4

And we need to take in count that the vertex form for a parabola is given by this expression:

y = a(x-h)^2 +k

By direct comparison we see that for this case:

a = 7, h = -5 and k=-4

And we know from the general expression that the vertex is:

V= (h,k)

So then the vertex for this case is:

V= (-5,-4)

4 0
3 years ago
3/7 % of a quantity is equal to what fraction of the quantity?
Mekhanik [1.2K]

Answer:

0.42

Step-by-step explanation:

0.42857142857

3 0
3 years ago
Read 2 more answers
The vertex of the parabola below is at the point (1,3), and the point (2,6) is
Inessa05 [86]

Answer:

B. y=3(x-1)2 + 3

Step-by-step explanation:

Given that

vertex of the parabola is at the point (1,3)

let's verify, if the option B is the correct equation of the parabola.

y=3(x-1)^2 + 3\\ \\y=3(x^2+1-2x) + 3\\\\y=3x^2+3-6x + 3\\\\y=3x^2-6x + 6....Eq1

comparing to standard equationof parabola (standard quadratic equation), we get

a=3, b=-6 and c=6

to find the vertex we use formula for x- coordinate as x=-b/2a

x=-(-6)/2(3)\\\\x=6/6\\x=1

to find y put x=1 in the Eq1, we get

y=3(1)^2-6(1)+6\\\\y=3-6+6\\\\y=3

vertex =(x,y) = (1, 3)

thus vertex of the parabola from the equation y=3(x-1)2 + 3 is (1,3), thus verified

7 0
3 years ago
1. The graph of y = x2 is translated 4 units down and 3 units to the left. It is also stretched by a factor
slega [8]

Answer:

ok so first ur gonna do 4x3=12 then

Step-by-step explanation:

7 0
2 years ago
Find all the zeroes of the polynomial function f(x)=x^3-5x^2+6x-30 using synthetic division.
WITCHER [35]
Write the coeeficientes of the polynomial in order:

  | 1   - 5   6   - 30
  |
  |
  |
------------------------

After some trials you probe with 5

   | 1   - 5       6     - 30
   |
   |
5 |        5      0       30
-----------------------------
     1     0      6        0 <---- residue

Given that the residue is 0, 5 is a root.

The quotient is x^2 + 6 = 0, which does not have a real root.

Therefore, 5 is the only root. You can prove it by solving the polynomial x^2 + 6 = 0.
3 0
3 years ago
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