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Natalija [7]
2 years ago
5

Find the real number root

Mathematics
1 answer:
Hunter-Best [27]2 years ago
7 0

Answer:

No real roots.

Step-by-step explanation:

Unless you guys are dealing with imaginary numbers, you cannot take the square root of a negative.

You might be interested in
Beth normally cycles a total distance of 56 miles per week.
Vikki [24]

Answer:

She will cycle 85.169 miles in the third week.

(round if needed)

Step-by-step explanation:

56÷100=0.56 (this is 1%)

0.56 x 15 + 56 = 64.4 (you find 15% then add 56 (the original amount) because it's an increase of 15%)

repeat for the other 3 weeks

64.4÷100=0.644

0.644 x 15 + 64.4 = 74.06

74.06÷100=0.7406

0.7406 x 15 + 74.06 = 85.169

8 0
2 years ago
In Algebra you learned that Difference of Perfect Squares is x^2 - 25 = (x+5)(x-5)
nika2105 [10]

Answer:

(x+5i)(x-5i)

Step-by-step explanation:

We use imaginary numbers to solve that. since i = sqare root of -1 so i * i = -1.

6 0
3 years ago
WHOEVER GETS IT RIGHT FIRST WILL BE CHOSEN THE BRAINLIEST!!
melisa1 [442]
The correct answer is x=12
8 0
2 years ago
Read 2 more answers
The Greatest Common Factor of 90,33,54
yawa3891 [41]
3 is the GCF of 903,354
3 0
3 years ago
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
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