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asambeis [7]
2 years ago
9

Helppppppppppppppppppppppppppppppppppppp

Mathematics
1 answer:
almond37 [142]2 years ago
4 0

i think it would be D

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Hich statement best describes a graph of paired points that form a proportional relationship?
Yanka [14]

Answer:

A. A straight line can be drawn through all the points and the line passes through the point (0,0) .

Step-by-step explanation:

See comment for options

Given

A graph with proportional relationship

Required

Which of options (a to d) describes the graph

A graph with a proportional relationship is represented as:

y = kx and

(x,y) = (0,0) is a paired point on the graph

This implies that options (b) and (c) are not true

Option (d) is also incorrect because if the graph does not pass through all points, then it is not proportional.

<em>Hence, (a) is correct</em>

6 0
3 years ago
Help please. i am confused
liraira [26]

Answer:

G(x+2) = 7x^2 + 33x + 30

Step-by-step explanation:

So in the G(x) function, to find G(x+2), we just simply plug in the value of x+2 into the function and the result is what is wanted. SO:

G(x+2) = 7(x+2)^2 + 5(x+2) -8 , which is 7x^2 +33x +30 after SIMP - lifying (see what I did there ;)

Hope i helped, please make this brainly. :)

4 0
3 years ago
Describe how the discriminant of a quadratic equation is related to the number of real solutions the equation has.
Anettt [7]
If the discriminant of a quadratic equation, given by B^2-4AC is positive, it has 2 real solutions. If 0, it has 1 real solution and if negative it has 0 real solutions.
5 0
3 years ago
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Numbers like 10 100 1000 and so on are called
Afina-wow [57]
Powers of ten! is the answer
4 0
4 years ago
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Find the value of a square + b square when a+b = 8 and ab = 10
Ne4ueva [31]

As we know,

\boxed{ \boxed{(a + b) {}^{2}  =  {a}^{2}  +  {b}^{2}  + 2ab}}

now, let's plug the given values to find a² + b²

  • (8) {}^{2}  =  {a}^{2}  +  {b}^{2}  + (2 \times 10)

  • 64 =   {a}^{2}  +  {b}^{2}  + 20

  • {a}^{2} +   {b}^{2}  = 64 - 20

  • a {}^{2} +  b {}^{2}  = 44
5 0
3 years ago
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