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Bad White [126]
2 years ago
14

The formula for distance is given as distance = speed x time. Use the formula to find:

Mathematics
1 answer:
goldenfox [79]2 years ago
3 0
You just put the numbers in the formula 3x30 = 90 km
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Cindy earns $10.50 per hour working at a restaurant. On Friday, she spent 1 3/4 hours cleaning, 2 1/3 hours doing paperwork, and
AnnyKZ [126]
$61.25

All you have to do is add all the fractions and multiply it by 10.50
7 0
3 years ago
WOULD YOU RATHER...
8090 [49]

Answer:

0.015 km/s

Step-by-step explanation:

40 km/hr

0.015 km/s

0.015 ÷ (1/3600)

0.015 × 3600

= 54 km/hr

7 0
3 years ago
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Is square meters a type of surface area
Irina18 [472]

Yes square meters is a type of surface area.

7 0
3 years ago
the endpoints of the diameter of a circle are (−6, 6) and (6, −2), what is the standard form equation of the circle?
miss Akunina [59]
The equation of a circle in standard form is

(x - h)^2 + (y - k)^2 = r^2

where (h, k) is the center of the circle, and r is the radius if the circle.

We need to find the radius and center of the circle.
We are given a diameter, so to find the center, we need the midpoint of the diameter.

M = ((-6 + 6)/2, (6 + (-2))/2) = (0, 2)
The center is (0, 2).

To find the radius, we find the length of the given diameter and divided by 2.

d = \sqrt{(-6 - 6)^2 + (6 - (-2))^2)}

d = \sqrt{144 + 64}

d = \sqrt{208}

r = \dfrac{d}{2} = \dfrac{\sqrt{208}}{2} = \dfrac{\sqrt{208}}{\sqrt{4}} = \sqrt{52}

(x - 0)^2 + (y - 2)^2 = (\sqrt{52})^2

x^2 + (y - 2)^2 = 52
7 0
3 years ago
Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En
ArbitrLikvidat [17]

Answer:

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, c) x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right), y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right).

Step-by-step explanation:

The equation of the circle is:

x^{2} + (y-1)^{2} = 16

After some algebraic and trigonometric handling:

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t

Where:

\frac{x}{4} = \cos t

\frac{y-1}{4} = \sin t

Finally,

x = 4\cdot \cos t

y = 1 + 4\cdot \sin t

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

c) x = 4\cdot \cos t'', y = 1 + 4\cdot \sin t''

Where:

4\cdot \cos t' = 0

1 + 4\cdot \sin t' = 5

The solution is t' = \frac{\pi}{2}

The parametric equations are:

x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right)

y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)

7 0
3 years ago
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