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arsen [322]
2 years ago
14

What is 0 raised to the second power?

Mathematics
1 answer:
astraxan [27]2 years ago
8 0

Answer:

0

Step-by-step explanation:

0 to the second power:

0x0= 0

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I'm marking answers as brainliest. ----------- What is the approximate solution to the system of equations? y=x+1, y=3x-2 ------
Alchen [17]

Answer:

Step-by-step explanation:

y = x + 1

y = 3x - 2

3x - 2 = x + 1

2x - 2 = 1

2x = 3

x = 3/2 or 1.5

y = 1.5 + 1

y = 2.5 or 2 1/2

(1.5, 2.5)

8 0
4 years ago
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A die is rolled 360 times. Let say that you want to use normal approximation to find the probability that the number of 4 was ro
Neporo4naja [7]

Answer:

0.0000

Step-by-step explanation:

Given that a die is rolled 360 times. Let say that you want to use normal approximation to find the probability that the number of 4 was rolled exactly 100 times.

Here we know that getting 4 is binomial with p = 1/6 and n =360

By approximating to normal after checking conditions are satisfied we have

X = no of fours obtained is normal

with mean = np =60

and variance = npq = 50

Std dev = 7.072

X is N(60, 7.072)

For finding out prob x=100 we have to do continuity correction as

P(X=100) = P(99.5

3 0
4 years ago
The radius of the large sphere is double the radius of the
Aleonysh [2.5K]

Answer:

8 times

Step-by-step explanation:

We know that the radius of smaller sphere is r,

The volume of sphere is given by:

V_1=\frac{4}{3} \pi r^{3}

where V_1 is the volume of the small sphere.

As we know that the radius of large sphere is double of the smaller sphere, the radius of large sphere will be 2r

Let V_2 be the volume of large sphere

V_2=\frac{4}{3}\pi (2r)^{3} \\ =\frac{4}{3}\pi *8r^3

Separating 8 aside

V_2=8(\frac{4}{3}\pi r^{3})\\V_2=8V_1

We can see that the volume of large sphere is eight times the volume of small sphere ..

7 0
3 years ago
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Please omg I need help ASAP please! (Giving brainliest)
Alla [95]

Answer:

Step-by-step explanation:

a 4x/5-3=12

4x/5=15

4x=75

x=75/4

b u could have done times 5 first

4x-3=60

3 0
3 years ago
Determine the length and conductivity of a wire with diameter 4.2 mm constructed from an alloy of resistivity 3.2 x 10^-8 ohms-m
xz_007 [3.2K]

Answer:

1) L = 3.465 × 10^2; 31250000 S/m

2) terminal pd = 11V

3) ₦2

Step-by-step explanation:

1)

diameter of wire (d) = 4.2mm = 4.2 × 10^-3m

resistivity 3.2 x 10^-8 ohms-meter

Resistance(R) = 0.8 ohms

Area of wire (A)= πd^2 / 4

A =[ ( 22/7) * (4.2×10^-3)^2] / 4

A = [22/7 * 17.64×10^-6] / 4

A = 13.86×10^-6m^2

Recall:

Resistivity = (R ×A) / Length

Length = (R × A) / resistivity

Length(L) = [(0.8)(13.86×10^-6)] / 3.2 x 10^-8

L = (11.088 × 10^-6) / 3.2 x 10^-8

L = 3.465 × 10^2

Conductivity = 1/resistivity

Conductivity = 1/3.2 x 10^-8

Conductivity = 31250000 S/m

2)

Series resistor = 1.5 and 4.0 ohms

e.m.f (E) = 12V, internal resistance 'r' = 0.5

Recall :

I = E / (R + r)

Where I = current

R = circuit resistance

R = Rs = 1.5 + 4.0 (series connection) = 5.5

Recall : I = V/R

Where V = voltage ( terminal pd)

Therefore,

V/R = E/(R+r)

V/5.5 = 12/(5.5 + 0.5)

V/5.5 = 12/6

V/5.5 = 2

V = 5.5 × 2

V = 11V

3) power = 100W, t =10 hours

Energy consumed = power × time

Energy consumed = 100W × 10 hrs = 1000Whr

Recall: 1 KW = 1000 W

Therefore, energy consumed = 1KW

Cost of energy per KW = ₦2

Therefore, cost of lightning the lamp = ₦2 × 1KW = $2

4 0
4 years ago
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