How do I make equivalent denominators for 1/5 and 2/10

Mathematics

2 answers:

LenaWriter [7]3 years ago

5 0

Multiply 5 by 2 and you'll get the equivalent denominator of 10.

(remember: what you do to the bottom you have to do to the top)
Hope it helps.

KonstantinChe [14]3 years ago

4 0

Find the Least Commmon Denominator =LCD then multiply what u get for the bottom ex: 2/8X3/24 multiply 8*3 for 2/8= 6/24

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What is the slope of the line passing through the points (−3, −5) and (−1, −6)

inessss [21]

slope = (y2-y1)/(x2-x1)

m= (-6--5)/(-1--3)

m = (-6+5)/(-1+3)

m=-1/2

4 0

3 years ago

Please help with the attached question. Thanks

Mademuasel [1]

Answer:

Choice A) $F(x) = 3\sqrt{x + 1}$ .

Step-by-step explanation:

What are the changes that would bring $G(x)$ to $F(x)$ ?

Translate $G(x)$ to the left by $1$ unit, and
Stretch $G(x)$ vertically (by a factor greater than $1$ .)

$G(x) = \sqrt{x}$ . The choices of $F(x)$ listed here are related to $G(x)$ :

Choice A) $F(x) = 3\;G(x+1)$ ;
Choice B) $F(x) = 3\;G(x-1)$ ;
Choice C) $F(x) = -3\;G(x+1)$ ;
Choice D) $F(x) = -3\;G(x-1)$ .

The expression in the braces (for example $x$ as in $G(x)$ ) is the independent variable.

To shift a function on a cartesian plane to the left by $a$ units, add $a$ to its independent variable. Think about how $(x-a)$ , which is to the left of $x$ , will yield the same function value.

Conversely, to shift a function on a cartesian plane to the right by $a$ units, subtract $a$ from its independent variable.

For example, $G(x+1)$ is $1$ unit to the left of $G(x)$ . Conversely, $G(x-1)$ is $1$ unit to the right of $G(x)$ . The new function is to the left of $G(x)$ . Meaning that $F(x)$ should should add $1$ to (rather than subtract $1$ from) the independent variable of $G(x)$ . That rules out choice B) and D).

Multiplying a function by a number that is greater than one will stretch its graph vertically.
Multiplying a function by a number that is between zero and one will compress its graph vertically.
Multiplying a function by a number that is between $-1$ and zero will flip its graph about the $x$ -axis. Doing so will also compress the graph vertically.
Multiplying a function by a number that is less than $-1$ will flip its graph about the $x$ -axis. Doing so will also stretch the graph vertically.

The graph of $G(x)$ is stretched vertically. However, similarly to the graph of this graph $G(x)$ , the graph of $F(x)$ increases as $x$ increases. In other words, the graph of $G(x)$ isn't flipped about the $x$ -axis. $G(x)$ should have been multiplied by a number that is greater than one. That rules out choice C) and D).

Overall, only choice A) meets the requirements.

Since the plot in the question also came with a couple of gridlines, see if the points $(x, y)$ 's that are on the graph of $F(x)$ fit into the expression $y = F(x) = 3\sqrt{x - 1}$ .