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Ket [755]
3 years ago
7

HELPPPPPP PLEASEEEEEEEEEEE

Mathematics
1 answer:
Minchanka [31]3 years ago
7 0
Using the power of power rule you
{ ({a}^{x}) }^{y}  =  {a}^{x \times y}
thus
({19}^{ \frac{1}{9} } )^9 =  {19}^{ \frac{1}{9} \times 9 }  =  {19}^{1}  = 19
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write the equation of a line that is parallel to y = - 3/2 x -1 and that passes through the point (4,6)​
maria [59]

Answer:

Step-by-step explanation:Given Equation of line has slope m=-3/2

So the line passing through (4,6) has the same slope because both are parallel

From slope intercept form we have

y-y1=m(x-x1)

y-6=-3/2(x-4)

y-6=-3/2x +3/2(4)

y=-3/2x+12

2y=-3x+24

3 0
3 years ago
Maya needs $58 to go on a field trip. She has saved $18.50. She earns $6.50 per hour cleaning her neighbor's garden, and she ear
insens350 [35]

She has $18.50 so she needs $39.50.

6.50×3= 19.50

18.50+19.50= 38

5.25×4=21

38+21=59

Maya would have enough money to go on the trip.

6 0
3 years ago
In the problem 4x + 10 list a term, factor, coefficient, and constant.​
kifflom [539]

term- 4x and 10

This is because to be a term it has to be + or -

Factor- 4x+10

This is because that is what is happening in the problem

Coefficient- 4y

I don't know what the constant would be, But I hope that this helps you

6 0
3 years ago
An adult brain is about 140 mm wide and divided into two sections (called "hemispheres" although the brain is not truly spherica
Sedbober [7]

Answer: 3.61×10^5 A

Step-by-step explanation: Since the brain has been modeled as a current carrying loop, we use the formulae for the magnetic field on a current carrying loop to get the current on the hemisphere of the brain.

The formulae is given below as

B = u×Ia²/2(x²+a²)^3/2

Where B = strength of magnetic field on the axis of a circular loop = 4.15T

u = permeability of free space = 1.256×10^-6 mkg/s²A²

I = current on loop =?

a = radius of loop.

Radius of loop is gotten as shown... Radius = diameter /2, but diameter = 65mm hence radius = 32.5mm = 32.5×10^-3 m = 3.25×10^-2m

x = distance of the sensor away from center of loop = 2.10 cm = 0.021m

By substituting the parameters into the formulae, we have that

4.15 = 1.256×10^-6 × I × (3.25×10^-2)²/2{(0.021²) + (3.25×10^-2)²}^3/2

4.15 = 13.2665 × 10^-10 × I/ 2( 0.00149725)^3/2

4.15 = 1.32665 ×10^-9 × I / 2( 0.000058)

4.15 × 2( 0.000058) = 1.32665 ×10^-9 × I

I = 4.15 × 2( 0.000058)/ 1.32665 ×10^-9

I = 4.80×10^-4 / 1.32665 ×10^-9

I = 3.61×10^5 A

3 0
3 years ago
V=pir^2h solve this equation for h
Nady [450]
Since we are solving for h, the equation:

V = πr²h

...divides πr² to the other side. Thus, it becomes....

\frac{V}{ \pi r^2} = h

Which can be rewritten as....

h = \frac{V}{ \pi r^2}
3 0
3 years ago
Read 2 more answers
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