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3 years ago

Please answer this correctly

Mathematics

2 answers:

harina [27]3 years ago

8 0

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maksim [4K]3 years ago

3 0

Answer:

3 minutes 56 minutes

Step-by-step explanation:

You might be interested in

A bottle contains 14/16 ounces of apple juice. One serving of apple juice weighs 1/4 ounce. What is the total number of servings

andrew-mc [135]

Answer:

You have 3.5 Servings.

Step-by-step explanation:

1/4 x 4 = 4/16

14/4 = 3.5

8 0

3 years ago

If m(x) =x+5/x-1 and n(x) =x-3 which function has the same domain as (m.n)(x)

Travka [436]

Answer:

Step-by-step explanation:

m (x) = (x+ 5) / (x -1) and n(x) = x - 3.

$m(x) = \dfrac{x+5}{x-1}$

Hence

$m(n) = \dfrac{n+5}{n-1}$

but n is a function and n = x-3. Now, replace value of n on m (n)

$(m(n))= \dfrac{(x-3)+5}{(x-3)-1}$

simplify it

$(m\bullet n)(x) =\dfrac{x+2}{x-4}\\$

since (m . n) (x) is a fraction function, its domain will be all x such that the denominator is NOT 0. That is

$x-4\ne 0\rightarrow x\ne 4$

among options, only C has the same denominator with $(m\bullet n)(x)$

Hence their domains are the same.

4 0

3 years ago

4 is added to three times a certain number the result is not more than when twice that number is added to 10 find the possible r

malfutka [58]

4
because
4*2+10= 18
4+4+4+4=16

3 0

3 years ago

What are the solutions to the equation

frosja888 [35]

Answer:

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

Step-by-step explanation:

You have the quadratic function $2x^2-x+1=0$ to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions $ax^2+bx+c=0$ with $a\neq 0$ the Bhaskara's Formula is:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}$

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

It usually has two solutions.

Then we have $2x^2-x+1=0$ where a=2, b=-1 and c=1. Applying the formula:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}$

Observation: $\sqrt{-1}=i$

$x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$

And,

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i$