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2 years ago

How how is 1/2 + 1/4 solve differently than 1/2 + 1/3?

Mathematics

2 answers:

yulyashka [42]2 years ago

7 0

1/2 + 1/4 =
2/4 + 1/4 =
3/4 <===

Vinil7 [7]2 years ago

6 0

Am not sure if there's a difference but I would solve them with using lcm method.

With the first question it can be easily determine as 2 is a factor of 4 however in the second question the lcm is found by multiplying the denominator.

$\frac{1}{2} + \frac{1}{4} \\ the \: lcm \: is \: 4\: \\ \frac{1 \times 2}{4} + \frac{1 \times 1}{4} = \frac{3}{4}$

$\frac{1}{2} + \frac{1}{3} \\ The \: lcm \: is \: 6 \\ \frac{1 \times 3}{6} + \frac{1 \times 2}{6} = \frac{5}{6}$

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2 years ago

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2 years ago

What are the domain and range of the function f(x) = 3x + 5?

AveGali [126]

I think the answer is B.

If the equation you mean is f(x)= 3x^2 +5.

Hope this helps!

4 0

2 years ago

Read 2 more answers

-4x-15y=-17<br>-x+5y=-13<br><img src="https://tex.z-dn.net/?f=%20-%204x%20-%2015y%20%3D%20%20-%2017%20%20%5C%5C%20%20-%20x%20%2B

salantis [7]

To solve this system of linear equations we need a little trick called elimination steps are below

we are given the following system of equations
$- 4x - 15y = - 17$
$- x + 5y = - 13$
if you look at the x variables in both equations you can see that we can easily eliminate them by multiplying the bottom equation by -4 as so
$- 4x - 15y = - 17$
$- 4( - x + 5y = - 13)$
simplify and we get
$- 4x - 15y = - 17$
$4x - 20y = 52$
now we can combine like terms on each side with the x's cancelling and get

$- 35y = 35$
now divide off -35
$y = - 1$
great now we can go back to our original system and pick a equation and substitute y back in to find x lets use

$- x + 5y = 13$
so now we substitute y and get
$- x + 5( - 1) = 13$
$- x - 5 = 13$
$- x = 17$
$x = - 17$

now we put x and y into one coordinate (x,y)
so now our FINAL ANSWER IS
(-17,-1)

6 0

2 years ago

PLEASE HELP

GrogVix [38]

Since the domain are the x-values of the function, we need to find the y-values using those x-values.
- for x= $\frac{1}{2}$ , $log x_{2} ( \frac{1}{2} )=-1$ . Now we have our first ordered pair: $( \frac{1}{2} ,-1)$
- for $x=1$ , $log_{2} (1)=0$ which gives us our next ordered pair: $(1,0)$
- for x=2, $log_{2} (2)=1$ , so (2,1)
- for x=4, $log_{2} (4)=2$ , so (4,2)
- for x=8, $log_{2} (8)=3$ , so our last order pair is (8,3)

Finally we can plug the points and joint them to create our graph:

3 0

2 years ago