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Assoli18 [71]
3 years ago
14

The number r is irrational. Which statement about r+7 is true? r+7 is rational r+7 is irrational r+7 can be rational or irration

al depending on the value of r
Mathematics
1 answer:
MrMuchimi3 years ago
6 0

Answer: r+7 is irrational

Step-by-step explanation:

1) If r is irrational, that means that it goes on and on forever.

2) 7 is a rational number because it can be expressed as a quotient of 2        integers (7÷1=7)

3) If you add a rational number to an irrational number it is still irrational. The numbers after the decimal point keep going and going. You are just changing the whole number, which if to the left of the decimal point.

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6(x^2-1).6x-1/6(x+1)
Andrej [43]
The answer is (<span><span>3.6<span>x^3 </span></span>− <span>3.766667x</span></span>+<span><span>−1/</span><span>6)

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7 0
3 years ago
Which of the following represents alpha decay?
Brilliant_brown [7]

The correct answer is the third option. I just got it correct.

6 0
4 years ago
Simplify and write the trigonometric expression in terms of sine and cosine:
asambeis [7]

Answer:

we have the expression as;

1/sin u cos u

Step-by-step explanation:

tan u = sin u/cos u

cot u = cos u/sin u

Thus;

sin u/cos u + cos u/sin u

The lcm is sin u cos u

Thus, we have that;

(sin^2 u + cos^2 u)/sin u cos u

But ; sin^2 u + cos^2 u = 1

so we have ;

1/sin u cos u

4 0
3 years ago
Three different non-zero digits can be arranged in six different ways to
alex41 [277]

Answer:

1134

Step-by-step explanation:

We have 3 digits:

a, b, c

a 3 digit number can be written as:

a*100 + b*10 + c*1

Such that these numbers can be:

{1, 2, 3, 4, 5, 6, 7, 8, 9}

Let's assume that:

a < b < c

Then the 3 smaller numbers are:

a*100 + b*10 + c

a*100 + c*10 + b

b*100 + a*10 + c

The 3 larger numbers are:

b*100 + c*10 + a

c*100 + a*10 + b

c*100 + b*10 + a

We know that the sum of the 3 smaller numbers is equal to 540, then:

(a*100 + b*10 + c) + (a*100 + c*10 + b) + (b*100 + a*10 + c) = 540

Let's simplify this:

(a + a + b)*100 + (b + c + a)*10 + (c + b + c) = 540

(2a + b)*100 + (b + c + a)*10 + (2c + b) = 540

The sum of the 3 larger numbers is equal to X, we want to find the value of X:

(b*100 + c*10 + a) + (c*100 + a*10 + b) + (c*100 + b*10 + a) = X

Now let's simplify the left side:

(b + c + c)*100 + (c + a + b)*10 + (a + b + a)*1 = X

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

Then we have two equations:

(2a + b)*100 + (b + c + a)*10 + (2c + b) = 540

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

Notice that the terms are inverted.

By looking at the first equation, we can see that:

(2c + b) = 10    (because the units digit of 540 is 0)

Then, we can see that:

(b + c + a + 1 ) = 14   (the one comes from the previous 10)

finally:

(2a + b + 1) = 5   (the one comes from the previous 14)

Then we can rewrite:

(2*c + b) = 10

(b + c + a) = 14 -1  = 13

(2a + b) = 5 - 1 = 4

Now we can replace these 3 in the equation:

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

(10)*100 + (13)*10 + 4 = X

1000 + 130 + 4 = X

1134 = X

The sum of the 3 largest numbers is 1134.

6 0
3 years ago
Please help me !! :(
dlinn [17]
22.8 sorry lf i am wrong
6 0
3 years ago
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