What does it mean when 6 people share 12 muffins

Which number rounds to 15,700,000 when rounded to the nearest hundred thousand?. A) 15,000,000. B) 15,579,999. C) 15,649,999. D)

Artyom0805 [142]

<span>The correct answer is option D. i.e. 15,659,999. Now, this number is closest to the given number i.e. 15,700,000. Becuase when 659 is rounded to the nearest number of hiher value then its value will be 700. Thus,15,659, 999 rounds to 15,700,000 when rounded to the nearest hundred thousand.</span>

6 0

3 years ago

What is the difference in minutes of median

REY [17]

<h2>Answer:</h2>

The answer is 25 minutes

<h2>Step-by-step explanation:</h2>

As we see the median time given in minutes before the installation was 53 minutes and after the installation it became 28 so by taking the difference which is 53 - 28 gives us 25 minutes.

8 0

3 years ago

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Julli [10]

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7 0

2 years ago

Find the standard equation of a sphere that has diameter with the end points given below. (3,-2,4) (7,12,4)

DiKsa [7]

Answer:

The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Let the first endpoint be represented as $(x_{1}, y_{1}, z_{1})$ and the second endpoint be $(x_{2}, y_{2}, z_{2})$ .

Hence,

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Midpoint = $(\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})$

Midpoint = $(\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\$

Midpoint = $(5, 5, 4)$

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

$d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }$

$d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}$

$d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}$

$d = \sqrt{16 +196 + 0$

$d =\sqrt{212}$

$d = 2\sqrt{53}$

This is the diameter

To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is $(5, 5, 4)$

Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

Hence, the equation of the sphere of radius $\sqrt{53}$ and center $(5, 5, 4)$ is

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}$

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

This is the standard equation of the sphere

6 0

3 years ago

Express the negations of each of these statements so that all negation symbols immediately precede predicates. a) ∀x∃y∀zT(x, y,

Gnesinka [82]

Answer:

a) ∀x∃y ¬∀zT(x, y, z)

∀x∃y ∃z ¬T(x, y, z)

b) ∀x¬[∃y (P(x, y) ∨ Q(x, y))]

∀x∀y ¬ [P(x, y) ∨ Q(x, y)]

∀x∀y [¬P(x, y) ^ ¬Q(x, y)]

c) ∀x ¬∃y (P(x, y) ^ ∃zR(x, y, z))

∀x ∀y ¬(P(x, y) ^ ∃zR(x, y, z))

∀x ∀y (¬P(x, y) v ¬∃zR(x, y, z))

∀x ∀y (¬P(x, y) v ∀z¬R(x, y, z))

d) ∀x¬∃y (P(x, y) → Q(x, y))

∀x∀y ¬(P(x, y) → Q(x, y))

∀x∀y (¬P(x, y) ^ Q(x, y))

3 0

3 years ago

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