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deff fn [24]
3 years ago
14

Which of the following is an example of the commutative property?

Mathematics
1 answer:
anyanavicka [17]3 years ago
6 0

Answer:

a(b+c)= ab + ac

Step-by-step explanation:

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The quotient of 10 plus x and y minus 3?
Ierofanga [76]
(10 + x) / (y - 3)....quotient is the result of division
5 0
3 years ago
Find the solution of y = –x – 3 for x = 1.
nikklg [1K]

Answer:

A.  :)

Step-by-step explanation:

5 0
3 years ago
A young woman wants to make at least $200 a week and can't work no more than 30 hours a week. she works at the library for $8 an
Bas_tet [7]
Let, Number of Hours in Library = x
Number of Hours in babysitting = y

Inequality would be: x + y = 30  ;
 &   8x + 6y ≥ 200

b) Solving: Multiply first equation by 6, 
6x + 6y ≥ 180
Subtract it from 2nd equation, 
2x ≥ 20
x ≥ 10

Now, substitute it in 1st equation, 
10 + y = 30
y = 20

Your Answer would be (10, 20)
x - value is lesser or equal to 10, so it would move to left (decreasing direction)!

Hope this helps!
5 0
3 years ago
How many gallons of pure alcohol should be added to 40 gallons of a 20% solution so
Rudik [331]

Answer:

8/3

Step-by-step explanation:

i put equations as this:

x+40 = y

where x is galllons of pure mixture and y is the gallons of the new mixture

the other equation i id was:

x+0.2(40) = 0.25y

x+8 = 0.25y

y = 4x+32

solve using substitution

(x+40) = 4x+32

x = 4x-8

-3x = -8

x = 8/3

3 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
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