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MAXImum [283]
3 years ago
5

Determine how many solutions exist for this system of equations:

Mathematics
1 answer:
IrinaK [193]3 years ago
4 0

Answer:

No Solution

Step-by-step explanation:

-3x+6=-3x-5

+5 +5

-3x+11=-3x

+3 +3

11=0

No Solution

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Please help my son asked me for this and I don't understand it.
Tasya [4]

Answer:

y = 2x + 3

Step-by-step explanation:

In the slope-intercept form of the equation of a line,

y = mx + b,

m = slope, and b = y-intercept.

Let's look at Adriana's equation and understand the parts:

y = 2x + 4

y = mx + b

m = slope = 2; b = y-intercept = 4

Now let's look at the description of Henry's equation.

He has the same slope as Adriana, so for Henry, m = 2 also.

His y-intercept is 1 less than Adriana's, so it is 1 less than 4. Henry's y-intercept is 3.

Now that we know that for Henry, m = 2, and b = 3, we can write his equation.

y = mx + b

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Answer: y = 2x + 3

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2 years ago
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qwelly [4]

Answer:

both the groups have the same value for the first quartile

7 0
3 years ago
Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -
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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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