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anyanavicka [17]
4 years ago
12

Which polynomials, given in square inches, could represent the area of a square with whole number side lengths if x is a whole n

umber greater than 2? Remember, the formula for the area of a square is A = s2. Check all that apply.
a) x2 − 9
b) x2 −100
c) x2 − 4x + 4
d) x2 + 10x + 25
e) x2 + 15x + 36
Mathematics
2 answers:
Natalija [7]4 years ago
5 0
X2 - 4x + 4 and x2 + 10x + 25 are both correct.
Deffense [45]4 years ago
5 0

Answer: Option c) and d) are correct.

Explanation: Since, we know that the area of a square, A= s^2, where,  s is the side of a square, and it always be a positive number and makes a perfect square.

In option (a),  x^2-9 does not gives a perfect square for all the values of x>2, where x is a whole number. So, it can not be the area of a square.

Similarly, option (b) and (e) can not give the perfect square for all the values of x greater than 2.

But, In option (c) and (d) equations make the perfect square.

Because, x^2+10x+25=0\Rightarrow (x+5)^2 and this gives a perfect square for every value of x. So, it can be the area of a square.

Similarly, x^2-4x+4=0\Rightarrow (x-2)^2 which also gives a perfect square for every value of x. So, it also can be the area of a square.

Thus, option (c) and (d) are the correct options.




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natta225 [31]
4 is the answer to the problem
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3 years ago
The Sky Ranch is a supplier of aircraft parts. Included in stock are 10 altimeters that are correctly calibrated and two that ar
mart [117]

Answer:  a) 0.545,b) 0.41, c) 0.045, d) not possible.

Step-by-step explanation:

The Sky Ranch is a supplier of aircraft parts. Included in stock are 10 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected without replacement. Let the random variable

x represent the number that are not correctly calibrated.

Complete the probability distribution table.

x={0,1,2,3}

P(x=0)=?

P(x=1)=?

P(x=2)=?

P(x=3)=?

Since we have given that

n = 12

number of good one = 3 from 10

number of bad one = 2

So, P(X=0)=\dfrac{^{10}C_3\times ^2C_0}{^{12}C_3}=\dfrac{120}{220}=0.545

P(X=1) = \dfrac{^{10}C_2\times ^2C_1}{^{12}C_3}=\dfrac{90}{220}=0.41

P(X=2)=\dfrac{^{10}C_1\times ^2C_2}{^{12}C_3}=\dfrac{10}{220}=0.045

P(X=3) is not possible.

Hence, a) 0.545,b) 0.41, c) 0.045, d) not possible.

5 0
3 years ago
Can this question be solved? 1/2x = 14​
Zielflug [23.3K]
X= 28 you just flip the 1/2 to 2/1 and multiply it on both sides
8 0
4 years ago
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Answer:

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4 0
3 years ago
Suppose that f (400 )equals3000 and f prime (400 )equals10. Estimate each of the following. ​(a) f (401 )​(b) f (400.5 )​(c) f (
disa [49]

Answer:

a) f(401) = 3010, b) f(400.5) = 3005, c) f(399) = 2990, d) f(398) = 2980, e) f(399.75) = 2997.5

Step-by-step explanation:

The estimation of each value can be found by the following value:

f(x + \Delta x) = f(x) + f'(x)\cdot \Delta x

a) f(401) = 3000 + 10\cdot (401-400)

f(401) = 3010

b) f(400.5) = 3000 + 10\cdot (400.5 - 400)

f(400.5) = 3005

c) f(399) = 3000 + 10\cdot (399 - 400)

f(399) = 2990

d) f(398) = 3000 + 10\cdot (398-400)

f(398) = 2980

e) f(399.75) = 3000 + 10\cdot (399.75-400)

f(399.75) = 2997.5

4 0
3 years ago
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