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3 years ago

What is the answer??

Mathematics

1 answer:

LekaFEV [45]3 years ago

7 0

Answer:

3 1/5 (or 3.5)

Step-by-step explanation:

2(7) - 7(-1)

_______

7 + -1

Next:

14 + 7 (two negatives make a positive)

_____

Next:

21/6 which equals 3 1/2 (3.5)

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Answer pls kkhejejrjt

balu736 [363]

Answer:

The answer is B :) hope you have a great day

7 0

4 years ago

If x>2, then x^2-x-6/x^2-4=

Viefleur [7K]

Consider the expression $\frac{x^{2} -x-6}{ x^{2} -4}$

To factorize the expression in the denominator we use difference of squares: $x^{2} -4=x^{2} - 2^{2} =(x-2)(x+2)$

To factorize $x^{2} -x-6$ we use the following method:

$x^{2} -x-6=(x-a)(x-b)$

where a, b are 2 numbers such that a+b= -1, the coefficient of x,

and a*b= -6, the constant.

such 2 numbers can be easily checked to be -3 and 2

(-3*2=6, -3+2=-1)

So $x^{2} -x-6=(x-a)(x-b)=(x+3)(x-2)$

$
 \frac{x^{2} -x-6}{ x^{2} -4}= \frac{(x+3)(x-2)}{(x-2)(x+2)}= \frac{x+3}{x+2}$

$\frac{x+3}{x+2}= \frac{x+2+1}{x+2}= \frac{x+2}{x+2}+ \frac{1}{x+2}=1+ \frac{1}{x+2}$

for x>2

$\frac{1}{x+2}\ \textless \ \frac{1}{2+2}= \frac{1}{4}$

thus

for x>2,

$1+ \frac{1}{x+2}\ \textless \ 1+ \frac{1}{4}= \frac{5}{4}$

Answer:

for x>2

$\frac{x^{2} -x-6}{ x^{2} -4} = \frac{x+3}{x+2} \ \textless \ \frac{5}{4}$ , (but the expression is never 0)

8 0

4 years ago

Help please!!!!!!!!!

faltersainse [42]

When the denominator = 0 and the numerator does not, you get a vertical asymptote. That said a = - 1

If the exponents on the highest variable are the same in the horizontal asymptote is the division of coefficient on the top with the bottom. In plain English what that means is that if you make m = 1 then the coefficients are 2 and 1 (2 in the numerator and 1 in the denominator). If m is any other value greater than 1 there is not a horizontal asymptote.

C <<<====answer.
Third one down.

8 0

3 years ago

Express the function F in the form f ∘ g. (Enter your answers as a comma-separated list. Use non-identity functions for f(x) and

damaskus [11]

Answer:

$F(x) = f \circ g$

$f(x) = x + 2$ and $g(x) = x$

Step-by-step explanation:

Given

$F(x) = x + 2$

Required

Express as

$F(x) = f \circ g(x)$

In functions:

$f \circ\ g(x) = f(g(x))$

So, we have:

$f(g(x)) = x + 2$

g(x) can be set to x and f(x) to x + 2

Hence:

$F(x) = f \circ g$

When: $f(x) = x + 2$ and $g(x) = x$

6 0

3 years ago

Which of the following has no solution?

Sholpan [36]

The first one.
This statement is saying that x is less than zero and greater than zero at the same time. This is not possible. A number cannot be both negative and positive.
It's not the second one because this one includes equal to zero. It can be lass than or equal to zero and greater that or equal to zero because it can be ZERO.
It's not the third because this one is an OR statement. It can be less than or equal to zero OR greater than or equal to zero.
I hope you understand

8 0

3 years ago