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Pavlova-9 [17]
3 years ago
10

Determine if the table represents a linear function. If so, find the equation that describes the function below?

Mathematics
1 answer:
oksian1 [2.3K]3 years ago
7 0
The correct answer is:  [D]:  "Table does not represent a linear function" .
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F(x)=2x+6
Klio2033 [76]

Answer:

Step-by-step explanation:

F(x) = 2x + 6

G(x) = - 5x - 9

product of F and G .....the product is the result of multiplication

(2x + 6)(-5x - 9) =

-10x^2 - 18x - 30x - 54 =

-10x^2 - 48x - 54 <====

6 0
3 years ago
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For what value of x is angle 1 complementary to angle 2, if m angle 1=5x+4 and m angle 2=7x+2?
belka [17]

<span>Two </span>Angles<span> <span>are </span></span>Complementary<span> <span>when they add up to 90 degrees (a Right </span></span>Angle). They don't have to be next to each other, just so long as the total is 90 degrees.

So (5x + 4) + ( 7x + 2) = 90

12x + 6 = 90

12x = 90 – 6

12x = 84

<span>X = 7</span>

6 0
3 years ago
What is the answer for x in the following equation<br> -0.1x + 5.43=-0.5x + 9.43
ELEN [110]
Not too sure if i can tell u much but the answer on my test is B
8 0
3 years ago
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Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -
GaryK [48]

Consider all parabolas:

1.

y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

2.

y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

3.

y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

4.

y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

5.

y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

6.

y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

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3 years ago
Find the missing exponent.
777dan777 [17]

Answer: what missing exponent?

Step-by-step explanation:

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