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3 years ago

9.4 yards to feet show conversion factors

1 answer:

6 0

There are 3 feet in 1 yard, that is 3ft/1yd or 1yd/3ft

you can use either, now, in this case, we want to convert
from yards to feet, so, that means we want to cancel out
the yards unit, so we use the form with the "yd" at the bottom
of the fraction, so it cancels out with the one above

$\bf 9.4\boxed{yd}\cdot \cfrac{3ft}{1\boxed{yd}}\implies \cfrac{9.4\cdot 3ft}{1}\implies 9.4\cdot 3ft$

You might be interested in

Check whether the relation R on the set S = {1, 2, 3} is an equivalent

kozerog [31]

Answer:

$R$ isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let $S$ denote a set of elements. $S \times S$ would denote the set of all ordered pairs of elements of $S\!$ .

For example, with $S = \lbrace 1,\, 2,\, 3 \rbrace$ , $(3,\, 2)$ and $(2,\, 3)$ are both members of $S \times S$ . However, $(3,\, 2) \ne (2,\, 3)$ because the pairs are ordered.

A relation $R$ on $S\!$ is a subset of $S \times S$ . For any two elements $a,\, b \in S$ , $a \sim b$ if and only if the ordered pair $(a,\, b)$ is in $R\!$ .

A relation $R$ on set $S$ is an equivalence relation if it satisfies the following:

Reflexivity: for any $a \in S$ , the relation $R$ needs to ensure that $a \sim a$ (that is: $(a,\, a) \in R$ .)

Symmetry: for any $a,\, b \in S$ , $a \sim b$ if and only if $b \sim a$ . In other words, either both $(a,\, b)$ and $(b,\, a)$ are in $R$ , or neither is in $R\!$ .

Transitivity: for any $a,\, b,\, c \in S$ , if $a \sim b$ and $b \sim c$ , then $a \sim c$ . In other words, if $(a,\, b)$ and $(b,\, c)$ are both in $R$ , then $(a,\, c)$ also needs to be in $R\!$ .

The relation $R$ (on $S = \lbrace 1,\, 2,\, 3 \rbrace$ ) in this question is indeed reflexive. $(1,\, 1)$ , $(2,\, 2)$ , and $(3,\, 3)$ (one pair for each element of $S$ ) are all elements of $R\!$ .

$R$ isn't symmetric. $(2,\, 3) \in R$ but $(3,\, 2) \not \in R$ (the pairs in $\! R$ are all ordered.) In other words, $3$ isn't equivalent to $2$ under $R\!$ even though $2 \sim 3$ .

Neither is $R$ transitive. $(3,\, 1) \in R$ and $(1,\, 2) \in R$ . However, $(3,\, 2) \not \in R$ . In other words, under relation $R\!$ , $3 \sim 1$ and $1 \sim 2$ does not imply $3 \sim 2$ .

3 0

3 years ago

Arnold's entire workout consisted of 10 minutes of warm up exercise, 25 minutes of lifting weights, and 15 minutes on the treadm

lord [1]

Add all the times together:

10 + 25 + 15 = 50 minutes total.

For the ratio divide the time for weights by total time:

25/50 which reduces to 1/2

The ratio is 1/2

5 0

3 years ago

Read 2 more answers

Drag the correct graph to the box under the equation it corresponds to.<br><br> HELP YOU GUYS

ra1l [238]

Answer:

<h2>See below</h2>

Step-by-step explanation:

I can't drag and drop the graphs, but I can graph the equations shown. Then, all you will have to do is match the graphs shown the the graph that I will provide.

<h3>EQUATION 1: y = x² - 2</h3>

Graph Properties:

Opens up

Vertex is 0, -2

Axis of symmetry is x = 0

Graph photo shown in file called equation 1 graph

<h3>EQUATION 2: y = 2x²</h3>

Graph Properties:

Opens up

Vertex is 0, 0

Axis of symmetry is x = 0

Graph photo shown in file called equation 2 graph

<h3>EQUATION 3: y = (x - 2)²</h3>

Graph Properties:

Opens up

Vertex is 2, 0

Axis of symmetry is x = 2

Graph photo shown in file called equation 3 graph

I'm always happy to help :)

5 0

3 years ago

Match each system to the correct choice.

pogonyaev

Answer:

Option A - Neither. Lines intersect but are not perpendicular. One Solution.

Option B - Lines are equivalent. Infinitely many solutions

Option C - Lines are perpendicular. Only one solution

Option D - Lines are parallel. No solution

Step-by-step explanation:

The slope equation is known as;

y = mx + c

Where m is slope and c is intercept.

Now, two lines are parallel if their slopes are equal.

Looking at the options;

Option D with y = 12x + 6 and y = 12x - 7 have the same slope of 12.

Thus,the lines are parrallel, no solution.

Two lines are perpendicular if the product of their slopes is -1. Option C is the one that falls into this category because -2/5 × 5/2 = - 1. Thus, lines here are perpendicular and have one solution.

Two lines are said to intersect but not perpendicular if they have different slopes but their products are not -1.

Option A falls into this category because - 9 ≠ 3/2 and their product is not -1.

Two lines are said to be equivalent with infinitely many solutions when their slopes and y-intercept are equal.

Option B falls into this category.

8 0

2 years ago

Is every terminating decimal an integer?? Yes or no

sweet [91]

Integers do not have decimals...so ur answer is no. terminating decimals are not integers

7 0

3 years ago