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frozen [14]
3 years ago
12

n an inequality, 0.5 is the greater number. What will the position of the other number be in relation to 0.5 on the vertical num

ber line?
Mathematics
2 answers:
sashaice [31]3 years ago
8 0

Answer:

Step-by-step explanation:

It is equal because they both are the same

Nadya [2.5K]3 years ago
4 0

Answer:

The other number would be located below 0.5 on the vertical number line.

Step-by-step explanation:

Hoped This helped!

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AlekseyPX

Answer:

1

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Step-by-step explanation: just cuz

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A student wrote a number pattern such that each number is 3 more that 4 times the previous number. What is the sixth term in the
Nesterboy [21]
Start with 1.
1st number = 1
2nd number = 1 * 4 + 3 = 7
3rd number = 7 * 4 + 3 = 31
4th = 31 * 4 + 3 = 127
5th = 127 * 4 + 3 = 511
Which means the 6th = 511 * 4 + 3, which is 2047.
4 0
3 years ago
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a carpebter has a board 2 meter needs to cut it in 8 equal pieces how many centerimeters long will each piece be
likoan [24]
2 centerimeters each
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3 years ago
Anitas a fast food chain specialising in hot dogs and garlic fries, keeps track of the proportion of its customers who decide to
ira [324]
<span>The probability that one customer will place a take-out order is .45.
The probability that one customer will place an "eat in" order is .55.

Of 4 randomly selected customers, the probability that exactly 3 will order "take out" is given by the following:
</span>C^4_3\times 0.45^3\times0.55^1\approx0.200475
6 0
3 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
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