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anygoal [31]
3 years ago
15

How many lengths of wire, each 3 3/4 inches. long can be cut from a 25-ft roll?

Mathematics
1 answer:
lianna [129]3 years ago
6 0

Answer:

80 pieces of wire.

Step-by-step explanation:

It may not come out evenly. If there is a remainder, it won't matter. Just throw it away and go with the integer answer.

Change 3 3/4 to a decimal

3.75

Change the 25 foot roll to inches.

25 feet * 12 inches / foot  = 300 inches.

Now how many pieces of wire do we get.

1 piece of wire/3.75 inches = x pieces of wire / 300 inches

3.75 * x = 300

3.75 * x / 3.75 = 300 / 3.75

x = 80

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Angle C is inscribed in circle O. Line AB is a diameter of circle O. What is the measure of angle B? I can't get the picture but
WINSTONCH [101]

Answer:

Step-by-step explanation:

Since the inscribed angle theorem tells us that any inscribed angle will be exactly half the measure of the central angle that subtends its arc, it follows that all inscribed angles sharing that arc will be half the measure of the same central angle. Therefore, the inscribed angles must all be congruent.

7 0
3 years ago
At what time the sum Rs. 5000 will be doubled at the interest<br>of 10% per annum?​
FromTheMoon [43]

Answer:

Time= 10 years

Step-by-step explanation:

Let total time=t

Initial Amount=5000

Final Amount=10000

Total interest = Final Amount - Initial Amount

                      =10000-5000\\=5000

Simple Interest =\frac{Initial Amount\times time\times rate}{100} \\\\5000=\frac{5000\times t \times 10}{100}\\\frac{t}{10} =1\\t=10 \ years

7 0
3 years ago
Which fraction expresses 36/80 in lowest terms? Please help me ;D
JulijaS [17]
To reduce<span> a </span>fraction to lowest terms<span> (also called its simplest form), divide both the numerator and denominator by the GCD. For example, 2/3 is in </span>lowest<span> form, but 4/6 is not in </span>lowest<span> form (the GCD of 4 and 6 is 2) and 4/6 can be expressed as 2/3.

38/80= 19/40
</span>
3 0
3 years ago
These roots of the polynomial equation x^4-4x^3-2x^2+12x+9=0 are 3,-1,-1. Explain why the fourth root must be a real number. Fin
Alex787 [66]

Roots with imaginary parts always occur in conjugate pairs. Three of the four roots are known and they are all real, which means the fourth root must also be real.

Because we know 3 and -1 (multiplicity 2) are both roots, the last root r is such that we can write

x^4-4x^3-2x^2+12x+9=(x-3)(x+1)^2(x-r)

There are a few ways we can go about finding r, but the easiest way would be to consider only the constant term in the expansion of the right hand side. We don't have to actually compute the expansion, because we know by properties of multiplication that the constant term will be (-3)(1)(1)(-r)=3r.

Meanwhile, on the left hand side, we see the constant term is supposed to be 9, which means we have

3r=9\implies r=3

so the missing root is 3.

Other things we could have tried that spring to mind:

- three rounds of division, dividing the quartic polynomial by (x-3), then by (x+1) twice, and noting that the remainder upon each division should be 0

- rational root theorem

4 0
3 years ago
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