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atroni [7]
2 years ago
14

Near a coral reef, a barracuda swam at a steady speed for 4.4seconds. It swam 33meters in that time. What was its speed?

Mathematics
1 answer:
PIT_PIT [208]2 years ago
3 0

Answer:

7.5 meters/second

Step-by-step explanation:

Speed = the magnitude of distance / time

Distance = 33 meters

Time = 4.4 seconds

Speed = 33 meters / 4.4 seconds = 7.5 meters/second. This is already to the tenths place, or the place right after the decimal, so we don't have to worry about rounding here.

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Please help me solve this question
Anastasy [175]

Answer:

50

Step-by-step explanation:

Use pythagoreom theorem to determine the length of the hypotenuse:

a² + b² = c²

- assign side a to 30 and side b to 40

30² + 40² = c²

90 + 160 = c²

250 = c²

Now square root both sides to isolate side c:

√250 = √c²

50 = c

side c = 50

7 0
3 years ago
(X2)3<br> Power of a power rule
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2•3=6
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2 years ago
The length of a rectangle is four times its width. If the perimeter of the rectangle is 20 feet, how many feet long is the recta
Gemiola [76]
Largeur = x
longueur = 4 x 

2 ( x + 4 x) = 20
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8 0
3 years ago
On a 4 1/2 hour trip Leslie drove 2/3 of the time for how many hours did Leslie drive
Greeley [361]

Answer:

3 hours

Step-by-step explanation:

4.5*(2/3)

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8 0
3 years ago
Read 2 more answers
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
2 years ago
Read 2 more answers
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