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3 years ago

Which is equivalent to V4500° after it has been simplified completely?

Mathematics

1 answer:

Rufina [12.5K]3 years ago

5 0

Answer:

Step-by-step explanation:

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I need help ASAP!!! Please explain how to solve the problem

azamat

Answer: The equation is $(x+4)^2 + (y-4)^2 = 9\\\\$

Center = (-4,4)

Radius = 3

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Explanation:

This graph shows the circle's center is at (-4, 4)

Recall in general, the center is (h,k)

So we see that (h,k) = (-4,4). More specifically: h = -4 and k = 4.

The radius of this circle is r = 3 as this is the distance from the center to any point on the edge. For example, go from (-4,4) to (-1,4).

With the h, k and r values in mind, we then say the following:

$(x-h)^2 + (y-k)^2 = r^2\\\\(x-(-4))^2 + (y-4)^2 = 3^2\\\\(x+4)^2 + (y-4)^2 = 9\\\\$

which represents the equation of this circle.

5 0

3 years ago

Need help!!! Due at 1:00!

LUCKY_DIMON [66]

Answer: D

Step-by-step explanation:

multiply the denominators then the numerators

7 0

4 years ago

On April 11, 2012, two earthquakes were measured off the northwest coast of Sumatra. The first had a magnitude of 8.6. The secon

Hoochie [10]

Answer:

Option D. 2.51 is correct.

Step-by-step explanation:

Magnitude of first earthquake = 8.6

Let intensity of first earthquake be x

$Magnitude = \log(\frac{x}{S}),\text{ where S is seismographic factor}\\\\\implies 8.6=\log(\frac{x}{S}).......(1)$

Magnitude of second earthquake = 8.2

Let intensity of second earthquake be y

$Magnitude = \log(\frac{y}{S}),\text{ where S is seismographic factor}\\\\\implies 8.2=\log(\frac{y}{S}).......(2)$

Now, to find the factor by which intensity of first earthquake is greater than intensity of second : we divide equation (1) by equation (2)

$\implies\log(\frac{x}{y})=8.6 - 8.2\\\\\implies \log(\frac{x}{y})=0.4\\\\\implies\frac{x}{y}=10^{0.4}\\\\\implies\frac{x}{y}=2.51$

Hence the required factor is 2.51

Therefore, Option D. 2.51 is correct.

4 0

3 years ago

This is a 2 part question.

irina [24]

B
just for extra space

4 0

3 years ago

Read 2 more answers

Find the length to the nearest centimeter of the diagonal of a square 30 cm. on a side

tiny-mole [99]

Answer:

42 cm.

Step-by-step explanation:

Please find the attachment.

Let x be the length of diagonal of the square.

We have been given that length of each side of a square is 30 cm. We are asked to find the length of the diagonal of square to the nearest centimeter.

We can see from our diagram that triangle AC is the diagonal of our square.

Since all the interior angles of a square are right angles or equal to 90 degrees, so we will use Pythagoras theorem to find the length of diagonal.

$AC^2=AD^2+DC^2$

Upon substituting our given values in above formula we will get,

$x^2=(30\text{ cm})^2+(30\text{ cm})^2$

$x^2=900\text{ cm}^2+900\text{ cm}^2$

$x^2=1800\text{ cm}^2$

Let us take square root of both sides of our equation.

$x=\sqrt{1800\text{ cm}^2}$

$x=42.4264\text{ cm}\approx 42\text{ cm}$

Therefore, the length of diagonal of our given square is 42 cm.

8 0

3 years ago