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IrinaK [193]
2 years ago
12

Please help! I'll give brainliest to whom is correct!

Mathematics
1 answer:
olga2289 [7]2 years ago
8 0

Answer:

I THINK IT IS THE LAST ONE

Step-by-step explanation:

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35 POINTS ANSWER FAST PLZ!!!!!!!!!!!!!!!
Sladkaya [172]

The formula to solve for thee area of a square pyramid is [ V = 1/3 * b² * h ]

b² = base area (b = length of one side of the base)

h = height

The only option that as a base length of 12cm. and a height of 14cm. is the first option.

Solve for the volume if needed.

V = 1/3(12²)(14)

V = 1/3(144)(14)

V = 1/3(2,016)

V = 672cm²

Best of Luck!

8 0
3 years ago
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Write the standard form of the equation of each line. <br> y=1/4 x+2
sweet-ann [11.9K]

y=mx+c where m = 1/4

c=2

3 0
3 years ago
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What is −30−6r= -36 ?
Nadya [2.5K]

-30-6r=36

We move all terms to the left:

-30-6r-(36)=0

We add all the numbers together, and all the variables

-6r-66=0

We move all terms containing r to the left, all other terms to the right

-6r=66

r=66/-6

r=-11

7 0
3 years ago
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The Lewis family is building a circular swimming pool if the radius of the pool is 9 feet what is it’s circumference use TT =3.1
zzz [600]

Answer:

C =56.52 ft

Step-by-step explanation:

The circumference is found by

C = 2*pi*r  where r is the radius

C = 2*3.14 * 9

C =56.52 ft

5 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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