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Sidana [21]
3 years ago
11

Write the coordinates of the point C after dilation with scale factor of 5, centered at the orgin

Mathematics
1 answer:
VladimirAG [237]3 years ago
4 0

Answer:

Coordinates of point C after dilation with a scale factor of 5, centered at the origin will be: C'(10, 5)

Step-by-step explanation:

Given the point C

  • C(2, 1)

A dilation with a scale factor of 5 means the size of the coordinates of the image of point P will be 5 times larger than than the original coordinates of point C.

Thus, the rule of dilation by a scale factor 5 is:

(x, y)  →  (5x, 5y)

C(2, 1)  →  (5x, 5y) = (5(2), 5(1)) = C'(10, 5)

Therefore, coordinates of the point C after dilation with a scale factor of 5, centered at the origin will be: C'(10, 5)

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Sindrei [870]
<h3>Solution:</h3>

<u>From the given data</u>

<u>93% of the bar is silver :</u>

\small\bold{ → 93\%    \: \: of \: \:   7kg}

\small \bold{→\frac{93}{100}  \times 7 }

\small \bold{→ \frac{651}{100}  }

\small \bold{→ 6.51 \: kg  }

<u>Therefore, 6.51 Kg of Silver is there in the bar.</u>

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2 years ago
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Tema [17]

Answer:

The density of the marble is 2,000 kg/m³

Step-by-step explanation:

The given parameters for finding the density of the marble are;

The volume of water in the graduated cylinder, v₁ = 20 ml

The mass of the marble dropped into the graduated cylinder, m = 8 g

The new volume of the cylinder after the 8 g marble is dropped into the cylinder, v₂ = 24 ml

Therefore, the volume of the marble, v = v₂ - v₁

Substituting the known values gives;

The volume of the marble, v = 24 ml - 20 ml = 4 ml

Density, ρ = Mass, m/(Volume, v), the units of density is kg/m³

Therefore, the density of the marble, \rho _{marble}, is given as follows;

\rho _{marble} = m/v = (8 g)/(4 ml) = 2 g/ml

1 g/ml = 1 kg/l = 1,000 kg/m³

∴ \rho _{marble} = 2 kg/l = 2,000 kg/m³.

4 0
3 years ago
What are the possible values of x in x2 + 3x + 3 = 0?
e-lub [12.9K]

To solve a quadratic equation like ax^2+bx+c=0, you can use the quadratic formula

x_{1,2} = \cfrac{-b\pm\sqrt{b^2-4ac}}{2a}

In your case, a = 1, b = c = 3, so the formula becomes

x_{1,2} = \cfrac{-3\pm\sqrt{(-3)^2-4\cdot 1\cdot 3}}{2\cdot 1}

We can simplify the expression:

x_{1,2} = \cfrac{-3\pm\sqrt{9-12}}{2} = \cfrac{-3\pm\sqrt{-3}}{2}

Since -3 is negative, its square root is computed as

\sqrt{-3} = \sqrt{-1\cdot 3} = \sqrt{-1}\sqrt{3} = \sqrt{3}i

So, the solutions are

x = \cfrac{-3+i\sqrt{3}}{2} \text{ or } x = \cfrac{-3-i\sqrt{3}}{2}

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Answer:

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Answer:

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