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Elza [17]
3 years ago
13

What should be subtracted from 3x to Gate 3 X - 2y​

Mathematics
1 answer:
castortr0y [4]3 years ago
3 0

Answer:

3x-2y

Step-by-step explanation:

2y must be subracted

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Okay so, i need help with this. It’s really important and I need to finish this paper before school tomorrow. Please help i will
SOVA2 [1]

Answer:

5x^2 - 4x - 3 = 0

Step-by-step explanation:

For the equation

ax^2 + bx + c = 0

the  roots are   [-b +/- √(b^2 - 4ac)] / 2a

So comparing the terms to the values in the question:

a = 5

b = -4

c = -3

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2 years ago
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7 0
3 years ago
N is the midpoint of AC ; AN = 4x - 7 and NC = 5x - 15<br> Find x=
xz_007 [3.2K]

Answer:

x = 8

Step-by-step explanation:

AN = NC,

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4 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
I will give brainlest if you get it correct ​
lianna [129]

Answer:i think d.please dont blame me im not sure.sorry

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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