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frosja888 [35]
2 years ago
9

Find the value of n, correct to 4 significant figures.​

Mathematics
1 answer:
Law Incorporation [45]2 years ago
7 0

Answer:

6.708

Step-by-step explanation:

  • Y varies directly as the square of x (Given)

  • \implies y\:\alpha \:x^2......(1)

  • Y varies inversely as the square root of z (Given)
  • \implies y\:\alpha\: \frac{1}{\sqrt z}......(2)

  • Combining (1) & (2), we find:

  • y\:\alpha \:\frac{x^2}{\sqrt z}

  • \implies y=\frac{kx^2}{\sqrt z} (Where k is constant of proportionality).....(3)

  • Now, when y = 2, x = 3 and z = 4, we find the value of k i.e. constant.

  • 2=\frac{k(3)^2}{\sqrt 4}

  • \implies 2=\frac{k(9)}{2}

  • \implies k =\frac{4}{9}

  • Plugging the value of k in (3), we find:

  • y=\frac{4x^2}{9\sqrt z} ....(4)

  • Next, in equation (4), plug y = 5, x = n and z = 16 and obtain the value of n by solving it.

  • 5=\frac{4(n)^2}{9\sqrt {16}}

  • \implies 5=\frac{4(n)^2}{9(4)}

  • \implies 5=\frac{(n)^2}{9}

  • \implies 5(9)=(n)^2

  • \implies 45=(n)^2

  • \implies n=\sqrt{45}

  • \implies n=6.70820393

  • \implies n\approx 6.708
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Answer:

48.6

Step-by-step explanation:

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Graph by using the slope and the y intercept.
Mashutka [201]

1. a) equation of the line :

  • y =  \dfrac{2}{5}x   - 7

y - intercept = -7

so, it will pass through point (0, -7)

and if we plug the value of x as 5, we get

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so, it will pass through point (5, -5) too

now, just plot the points (0 , -7) and (5 , -5) and join them.

2. b) equation of line is :

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here, y - intercept = 5

so the line passes through point (0 , 5)

now, Plugging the value of x = 1 we get :

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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.

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

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3 years ago
Please answer see the image
mash [69]

Answer:

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Step-by-step explanation:

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               = 67.7

8 0
3 years ago
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