2 years ago

(5+6)-+(-5-6)+-(5--6)

Mathematics

2 answers:

cupoosta [38]2 years ago

8 0

Answer:

Step-by-step explanation:

5+6=11

-5-6= -11

5--6=11

11+11=22

22-11=11

answer=11

stellarik [79]2 years ago

7 0

Answer:

$(5 + 6) + (5 - 6) + (5 - 6) \\ 11 + ( - 1) + ( - 1) \\ 11 - 1 - 1 \\ 10 - 1 \\ 9$

You might be interested in

(1, 5) (3 , 1) <br> Please Help

hammer [34]

Answer:

(4, 6)?

Step-by-step explanation:

Im not sure you didn't give much info

6 0

2 years ago

Given the functions, f(x) = x2 + 2 and g(x) = 4x - 1, perform the indicated operation. When applicable, state the domain restric

icang [17]

I hope this helps you

7 0

3 years ago

Which of the following decimals is equivalent to three and ten hundredths

Montano1993 [528]

Answer: 3.1 or 3.10

Step-by-step explanation:

3 and 10 hundredths = 3 + 10/100 = 3 + 0.10

3.10 / 3.1

6 0

3 years ago

prove that if f is a continuous and positive function on [0,1], there exists δ > 0 such that f(x) > δ for any x E [0,1] g

ValentinkaMS [17]

Answer:

I dont Know

Step-by-step explanation:

5 0

3 years ago

Lim x-1 x2 - 1/ sin(x-2)

balu736 [363]

Answer:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0$

Explanation:

Assuming the correct expression is to find the following limit:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}$

Use the property the limit of the quotient is the quotient of the limits:

$\lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}$

Evaluate the numerator:

$\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}$

Evaluate the denominator:

Since $\lim_{x \to1}sin(x-2)\neq 0$

$\frac{0}{\lim_{x \to1}sin(x-2)}=0$

4 0

3 years ago