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HACTEHA [7]
3 years ago
13

Solve 2(x-1) =4 xxxxxx

Mathematics
1 answer:
sineoko [7]3 years ago
4 0
2(x-1)=4
2x-2=4
2x=6
x=3
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In rectangle KLMN, KM = 10x+24 and KN = 64 find x
pishuonlain [190]
Hi there, I don't know if I am right but I will try to solve this, 10x+24=64, subtract 24 from both sides, 10x+24-24=64-24=10x=40, divide both sides by 10, 10x/10=40/10, x=4
7 0
3 years ago
Solve the equation square root of x plus three plus four equals five for the variable. Show each step of your solution process
alekssr [168]

Answer:

x=-2

Step-by-step explanation:

The given expression is

\sqrt{x+3}+4=5

We group the constant terms on the right hand side to obtain;

\sqrt{x+3}=5-4

\Rightarrow \sqrt{x+3}=1

We square both sides of the equation to remove the square root.

\Rightarrow (\sqrt{x+3})^2=1^2

x+3=1

This will simplify to;

x=1-3

x=-2

4 0
3 years ago
Read 2 more answers
For g(x)=x^2-2, determine g(x+2)
Svetlanka [38]

Answer:g(x)=-4^2+4x-5

Step-by-step explanation:

6 0
2 years ago
Every week at the Cougar Scouts meeting, Diane is in charge of the craft activity. This week, she decides to make pinecone bird
liraira [26]

Answer:

I AM NOT ABLE TO GIVE THE ANSWER OF THIS QUESTION

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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