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SpyIntel [72]
3 years ago
12

An alligator can run at a speed of 13 ft./s on land at this rate how far can it run in three seconds ?

Mathematics
1 answer:
SOVA2 [1]3 years ago
3 0
If it can run 13 feet in one second, then multiply 13 times 3. He will run 39 feet in 3 seconds.
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Rewrite the standard form equation below into slope-intercept form.<br>x – 2y = 3
Gnom [1K]
X-2y = 3

-2y=3-x move the x over to the other side 

-y=3/2 -1/2x divide both sides by 2

y=1/2x - 3/2 divide both sides by -1

8 0
3 years ago
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How many miles from ft Lauderdale to Haiti
Sonbull [250]

Ft. Lauderdale is a city that stretches several miles, and Haiti
is a whole country, that's many miles wide and many miles high. 

In order to nail down a reliable answer, you'd really need to specify
one point in Ft. Lauderdale and one point in Haiti.

If you start at the northwest end of Runway-13 at Ft. Lauderdale
Executive Airport, and take the shortest possible route to the east
end of Runway-28 at the Aeroport International de Port au Prince
at Haiti's capital city, you'd have to travel 727.57 miles.

But if you start in Ft. Lauderdale at the intersection of Griffin Rd
and Ravenswood Rd, and take the shortest possible route to the
Dispensaire de Bord-de-Mer hospital on Haiti's north coast, you'd
only have to travel 613.63 miles.

You really need to say WHERE in Ft. Lauderdale and WHERE in Haiti.
3 0
3 years ago
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Identify the Pythagorean triple.
Phantasy [73]

Answer:

I'm almost positive it's .A but I'm not sure

4 0
3 years ago
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a. 0.98 – 0.053 b. 0.67 – 0.4 c. 0.3 – 0.002 d. 3.2 – .789 e. 6.53 – 4.298 f. 6 – 4.32 g. 7 – 3.574 h. 4.83 – 1.8 i. 3.7 – 1.8 j
Pavlova-9 [17]

a. 0.927

We have:

0.98 – 0.053

We can re-write it as:

0. 9 8 0 -

0. 0 5 3

Moving digits to the right:

0. 9 7 10 -

0. 0 5 3

Digit-per-digit subtraction:

0. 9 2 7


b. 0.27

We have:

0.67 – 0.4

We can re-write it as:

0. 6 7 -

0. 4 0

Digit-per-digit subtraction:

0. 2 7


c. 0.298

We have:

0.3 – 0.002

We can re-write it as:

0. 3 0 0 -

0. 0 0 2

Moving digits to the right:

0. 2 10 0 -

0. 0 0 2

Again:

0. 2 9 10 -

0. 0 0 2

Digit-per-digit subtraction:

0. 2 9 8


d. 2.411

We have:

3.2 – .789

We can re-write it as:

3. 2 0 0 -

0. 7 8 9

We need to rewrite the first term by moving digits to the right several times:

3. 2 0 0 = 2. 12 0 0 = 2. 11 10 0 = 2. 11 9 10

So now we have:

2. 11 9 10 -

0. 7 8 9

Digit-per-digit subtraction:

2. 4 1 1


e. 2.232

We have:

6.53 – 4.298

We can re-write it as:

6. 5 3 0 -

4. 2 9 8

We need to rewrite the first term by moving digits to the right several times:

6. 5 3 0 = 6. 5 2 10 = 6. 4 12 10

So now we have:

6. 4 12 10 -

4. 2 9 8

Digit-per-digit subtraction:

2. 2 3 2


f. 1.68

We have:

6 – 4.32

We can re-write it as:

6. 0 0 -

4. 3 2

We need to rewrite the first term by moving digits to the right several times:

6. 0 0 = 5. 10 0 = 5. 9 10

So now we have:

5. 9 10 -

4. 3 2

Digit-per-digit subtraction:

1. 6 8


g. 4.426

We have:

7 – 3.574

We can re-write it as:

7. 0 0 0 -

3. 5 7 4

We need to rewrite the first term by moving digits to the right several times:

7. 0 0 0 = 6. 10 0 0 = 6. 9 10 0 = 6. 9 9 10

So now we have:

6. 9 9 10 -

3. 5 7 4 =

Digit-per-digit subtraction:

3. 4 2 6


h. 3.03

We have:

4.83 – 1.8

We can re-write it as:

4. 8 3 -

1. 8 0

We can immediately do the digit-per-digit subtraction:

3. 0 3


i. 2.9

We have:

3.7 – 1.8

We can re-write it as:

3. 7 -

1. 8

We need to rewrite the first term by moving digits to the right:

3. 7 = 2. 17

So now we have:

2. 17 -

1. 8 =

Digit-per-digit subtraction:

2. 9


j. 4.538

We have:

16.17 – 11.632

We can re-write it as:

1 6 . 1 7 0 -

1 1 . 6 3 2

We need to rewrite the first term by moving digits to the right:

1 6. 1 7 0 = 1 6. 1 6 10 = 1 5. 11 6 10  

So now we have:

1 5. 11 6 10 -

1 1. 6 3 2 =

Digit-per-digit subtraction:

0 4. 5 3 8

3 0
3 years ago
Find the distance traveled by a particle with position (x, y as t varies in the given time interval. x = 3?sin2 t, y = 3?cos2 t,
ANTONII [103]
Assuming the particle's position is given by

\begin{cases}x(t)=3-\sin2t\\y(t)=3-\cos2t\\0\le t\le4\end{cases}

then the distance traveled over the interval is

\displaystyle\int_{t=0}^{t=4}\sqrt{\left(\dfrac{\mathrm dx(t)}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy(t)}{\mathrm dt}\right)^2}\,\mathrm dt
=\displaystyle\int_0^4\sqrt{(-2\cos2t)^2+(2\sin2t)^2}\,\mathrm dt
=\displaystyle\int_0^4\sqrt{4\cos^22t+4\sin^22t}\,\mathrm dt
=\displaystyle2\int_0^4\mathrm dt
=2(4-0)=8
5 0
3 years ago
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