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kumpel [21]
3 years ago
11

This is worth 32 points and Ill mark brainlyess whoever gets it right

Mathematics
1 answer:
Valentin [98]3 years ago
4 0
The answer is less than.....too sure
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Find an equation for the line that passes through the given point and satisfies the given conditions. 11) 11) p = ( 9 , 2 ); per
geniusboy [140]

You can use the direction vector as the coefficients of x and y for a line through (0, 0) perpendicular to that direction: x + y = 0. (A common factor of 4 can be removed from the coefficients.)

Translating the line up 2 and right 9, so it goes through the given point, we get ...

... (x -9) +(y -2) = 0

or

... x + y = 11

5 0
3 years ago
The table shows the maximum and minimum depths of two submarines. Find the range of depths for each submarine. Then determine wh
JulsSmile [24]

Answer:

Range A = 386ft

Range B = 427ft

B has the larger range.

Explanation:

To find the range, find the difference between the minimum and maximum depths for each submarine. Remember that your answer will be positive.

Sub A:

-146ft - (-532ft) = 386 ft

Sub B:

-194ft - (-621ft) = 427 ft

Submarine B has a larger range.

7 0
3 years ago
74 ÷ 2072= <br> PLEASE HELPPPP
Anon25 [30]

Answer:

0.0357142857142857

8 0
3 years ago
Read 2 more answers
CAN SOMEONE WHO KNOWS HOW TO DO THIS PLEASE HELP ME
7nadin3 [17]
Firstly start off with finding the lowest common multiple of 14 and 24. The LCM would be 168. Then you would look at your powers, here you have to choose the highest power of each letter. So 14x^4 and 24x^6 the highest is x^6, then for y it would be y^6 and then for z it would be z^8. So altogether your answer would be 168x^6y^6z^8
4 0
3 years ago
Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is bein
ivanzaharov [21]

Answer:

1508527.582 cm³/min

Step-by-step explanation:

The net rate of flow dV/dt = flow rate in - flow rate out

Let flow rate in = k. Since flow rate out = 6800 cm³/min,

dV/dt = k - 6800

Now, the volume of a cone V = πr²h/3 where r = radius of cone and h = height of cone

dV/dt = d(πr²h/3)/dt = (πr²dh/dt)/3 + 2πrhdr/dt (since dr/dt is not given we assume it is zero)

So, dV/dt = (πr²dh/dt)/3

Let h = height of tank = 12 m, r = radius of tank = diameter/2 = 3/2 = 1.5 m, h' = height when water level is rising at a rate of 21 cm/min = 3.5 m and r' = radius when water level is rising at a rate of 21 cm/min

Now, by similar triangles, h/r = h'/r'

r' = h'r/h = 3.5 m × 1.5 m/12 m = 5.25 m²/12 m = 2.625 m = 262.5 cm

Since the rate at which the water level is rising is dh/dt = 21 cm/min, and the radius at that point is r' = 262.5 cm.

The net rate of increase of water is dV/dt = (πr'²dh/dt)/3

dV/dt = (π(262.5 cm)² × 21 cm/min)/3

dV/dt = (π(68906.25 cm²) × 21 cm/min)/3

dV/dt = 1447031.25π/3 cm³

dV/dt = 4545982.745/3 cm³

dV/dt = 1515327.582 cm³/min

Since dV/dt = k - 6800 cm³/min

k = dV/dt - 6800 cm³/min

k = 1515327.582 cm³/min - 6800 cm³/min

k = 1508527.582 cm³/min

So, the rate at which water is pumped in is 1508527.582 cm³/min

5 0
3 years ago
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