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Likurg_2 [28]
3 years ago
9

Find the slope of (14, 8) and (7, -6)

Mathematics
1 answer:
maksim [4K]3 years ago
5 0

Answer:

The Slope = 2

Explanation:

Slope : Δy/Δx = 2/1 = 2

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The zeros of this using quadratic equation​
aleksandr82 [10.1K]

Answer:

you have to use the quadratic equation to find the zeros

Step-by-step explanation:

3 0
4 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 = -
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.

3 0
3 years ago
What is p?<br><br> 7 1/5+p+4 9/10+3 3/20=19 1/10
Yakvenalex [24]
ANSWER:
7 \frac{1}{5}+p+4 \frac{9}{10}+3 \frac{3}{20} =19 \frac{1}{10} \\ 7 \frac{4}{20}+p+4 \frac{18}{20} + 3 \frac{3}{20}=19 \frac{2}{20}  \\ p=19 \frac{2}{20}- (7 \frac{4}{20}+4 \frac{18}{20} + 3 \frac{3}{20})  \\ p=19 \frac{2}{20}-15 \frac{5}{20} \\ p=3\frac{17}{20}  
6 0
3 years ago
What is the solution for the equation?
Vesnalui [34]

Answer:

The soultion is the answer.

Step-by-step explanation:

3 0
3 years ago
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Solve. Find all solutions in [0,2).5 secx cotx+5 secx+ cotx+1=0
Dmitrij [34]
\begin{gathered} 5sec(x)cot(x)+5sec(x)+cot(x)+1=0 \\ Factoring \\ (cot(x)+1)(5sec(x)+1)=0 \\ cot(x)+1=0 \\ cot(x)=-1 \\ x=cot^{-1}(1) \\ x=\frac{3\pi}{4} \\  \\ 5sec(x)+1=0 \\ 5sec(x)=-1 \\ sec(x)=\frac{-1}{5},\text{ there is no solution} \\ Hence \\ All\text{ solution are }\frac{3\pi}{4}+\pi n,\text{ where n=1,2,3...} \\  \\  \end{gathered}

5 0
1 year ago
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