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3 years ago

What is 3 1/2 - 1 5/9

Mathematics

2 answers:

jek_recluse [69]3 years ago

8 0

Answer: 2 1/18

Step-by-step explanation:

Sergeu [11.5K]3 years ago

5 0

Answer:

35/18

Step-by-step explanation:

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2x - 5y +10x - 2y + 3z solve.

Margarita [4]

Answer:

12x-7y+3z

Step-by-step explanation:

8 0

3 years ago

simplify the following expression: (x+7/4x) + (5/x-8) I also attached a photo of the problem if you get confused on how I typed

Gnoma [55]

Answer:

$\frac{x^{2} + 19x - 56}{4 {x}^{2} - 32x}$

Step-by-step explanation:

Start by making the denominators of both fraction the same. This can be done by multiplying one fraction's denominator by the other.

After simplifying and combining the two fractions together, check if the numerator can be factorised such that there is a common factor in the denominator and numerator. In this case, the numerator cannot be factorised.

Lastly, expand the denominator.

$\frac{x + 7}{4x} + \frac{5}{x - 8}$

$= \frac{(x + 7)(x - 8)}{4x(x - 8)} + \frac{5(4x)}{4x(x - 8)}$

$= \frac{x(x) + x( - 8) + 7(x) + 7( - 8) + 20x}{4x(x - 8)}$

$= \frac{ {x}^{2} - 8x + 7x - 56 + 20x }{4x(x - 8)}$

$= \frac{ {x}^{2} + 19x - 56}{4x(x - 8)}$

$= \frac{ {x}^{2} + 19x - 56}{4 {x}^{2} - 32x}$

7 0

2 years ago

The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .