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Anna [14]
3 years ago
10

a school bought 831 boxes of computer paper for the computer lab. Each box has 59 sheets of paper inside it. how many sheets of

paper were bought in total?
Mathematics
2 answers:
Step2247 [10]3 years ago
7 0
Hey there!

All you have to do is simply multiply!

831*59= 49,029

49,029 total sheets of paper were bought
Bess [88]3 years ago
4 0
<span> 831 *59=49,029

hope this helps!</span>
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A rectangle has a perimeter of 51.6 feet and a height of 9.9 feet. What is the length of the base?
beks73 [17]

Rectangle:

P = 2(L + W)

51.6 = 2 (L + 9.9)

L + 9.9 = 51.6 / 2

L + 9.9 = 25.8

L = 15.9


Answer

The length of the base = 15.9 feet

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3 years ago
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Aleonysh [2.5K]

Answer: The answer is 36

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Evaluate the expression for the given value of the variable(s).<br><br> -x2 - 4x; x = –3
yaroslaw [1]

Answer:

3

Step-by-step explanation:

Subbing -3 for x in -x^2 - 4x, we get -(-3)^2 - 4(-3)

                                                       OR -9       +12

Final result:  3

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3 years ago
You decide to purchase 30 toys, Your first few customers each buy 3 toys, and
grigory [225]

Answer:

30- 3x \leq 14

Step-by-step explanation:

Let say that Y is the number of toys you have now and X is the number of the customer. Every customer (X) buy 3 toys, and you have 30 toys at the start. So you can put this equation

30- 3x = Y

Now you have at least 14 toys, in other words, it is more than or equal to 14 (\leq14). The equation will be:

30- 3x \leq 14

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
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