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cestrela7 [59]
3 years ago
14

Really could use the help

Mathematics
1 answer:
SVEN [57.7K]3 years ago
7 0
c squared= a squared+b squared so c squared=25+64 so the answer would be the square root of 89. Hope this helps!!
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PLEASEEEE HELP!! This is my SECOND time asking :((
adoni [48]

Answer:

C

Step-by-step explanation:

formula: <u>Y</u><u>2</u><u> </u><u>-</u><u> </u><u>Y</u><u>1</u>

X2 - X1

-18 = X1

16 = Y1

31 = X2

40 = Y2

<u>4</u><u>0</u><u> </u><u>-</u><u> </u><u>1</u><u>6</u><u>. </u><u> </u><u> </u> = <u>2</u><u>4</u>

31 - (-18) 49

3 0
3 years ago
The height of the water in a tank decreases 3 inches each week due to evaporation and every two weeks a man adds 5 inches of wat
atroni [7]
75

Reduction of 12
3 x 4 = 12

Increase of 10
5 x 2 = 10

75 - 12 = 63
63 + 10 = 73

Change = 2 inches (75 - 73)
7 0
2 years ago
Read 2 more answers
Please help with this problem!
Dovator [93]

Answer:

x+14

Step-by-step explanation:

(12+x)+2

12+x+2

x+(12+2)

x+14

4 0
3 years ago
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Boating across a river, the current moves you downstream. Instead of going directly across the river, the boat moves at a 25 deg
solong [7]
If it's described as my drawing:

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7 0
3 years ago
A submarine left Diego Garcia five hours before a cruise ship. The ships traveled in opposite directions. The cruise ship travel
Rina8888 [55]

Answer:

The speed of the submarine is 15.429 miles per hour.

Step-by-step explanation:

Let suppose that both ships travel at constant velocities. As we know that both travel in opposite directions, it is supposed that cruise ship moves in +x direction, whereas submarine in -x direction. Kinematic equations for each sheep are described below:

Ship

x_{Sh} = x_{o} + v_{Sh}\cdot t

Submarine

x_{Su} = x_{o}+v_{Su}\cdot t

Where:

x_{o} - Position of Diego Garcia island, measured in miles.

x_{Sh}, x_{Su} - Current positions of ship and submarine, measured in miles.

v_{Sh}, v_{Su} - Velocities of ship and submarine, measured in miles per hour.

t - TIme, measured in hours.

If we know that x_{Sh} - x_{Su} = 241\,mi, v_{Sh} = 19\,\frac{mi}{h} and t = 7\,h, then:

x_{Sh} - x_{Su} = (v_{Sh}-v_{Su})\cdot t

We clear now the velocity of submarine:

\frac{x_{Sh}-x_{Su}}{t} = v_{Sh}-v_{Su}

v_{Su} = v_{Sh}-\frac{x_{Sh}-x_{Su}}{t}

v_{Su} = 19\,\frac{mi}{h} -\frac{241\,mi}{7\,h}

v_{Su} = -15.429\,\frac{mi}{h}

Speed of the submarine is the magnitude of its velocity, which is 15.429 miles per hour.

4 0
3 years ago
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