All categories

2 years ago

Solve the equation using the distributive property and properties of equality.

Mathematics

2 answers:

RideAnS [48]2 years ago

6 0

7 1/2 ummm i did it in my head lol but i’m sure it’s right

Inessa [10]2 years ago

6 0

The answer Is def 7 1/2

You might be interested in

Express in the form of p/q :<br><br> (i) 0.12▁3 (ii) 0. ▁621 (iii) 2.2▁18

konstantin123 [22]

Answer:

1)0.123( bar on 3)

Let X = 0.123 ( bar on 3)

Then X = 0.1233 --------(1)

Multiply equation (1) by 100 we get,

100X = 12.33 --------(2)

Again multiply equation (2) by 10 we get,

1000X = 123.33 -------(3)

Subtract equation (2) from equation (3) we get,

1000X = 123.33

100X = 12.33

____________

900X = 111

X = 111/900

Hence,

0.123 ( bar on 3) is in the form of 111/900 which is in the form of P/Q.

6 0

2 years ago

Find all integers $n$ for which $\frac{n^2+n+1}{n-1}$ is an integer.

Grace [21]

$\dfrac{n^2+n+1}{n-1}=\dfrac{(n-1)^2+3(n-1)+3}{n-1}=n-2+\dfrac3{n-1}$

The remainder term $\dfrac3{n-1}$ will never vanish, and will always be rational as long as the denominator is not 1, 3, or -3.

$n-1=1\implies n=2$
$n-1=3\implies n=4$
$n-1=-3\implies n=-2$

7 0

2 years ago

From the set {20, 38, 47}, use substitution to determine which value of x makes the inequality true.

vredina [299]

One way is to solve (it's easier)

x-9>29
add 9
x+9-9>29+9
x+0>38
x>38
just pick the numbers bigger than 38 as answer

47 is only number (38>38 is false)

answer is A

7 0

2 years ago

Suppose you have 5 riders and 5 horses, and you want to pair them off so that every rider is assigned one horse (and no horse is

maw [93]

There are 120 ways in which 5 riders and 5 horses can be arranged.

We have,

5 riders and 5 horses,

Now,

We know that,

Now,

Using the arrangement formula of Permutation,

i.e.

The total number of ways $^nN_r = \frac{n!}{(n-r)!}$ ,

So,

For n = 5,

And,

r = 5

As we have,

n = r,

So,

Now,

Using the above-mentioned formula of arrangement,

i.e.

The total number of ways $^nN_r = \frac{n!}{(n-r)!}$ ,

Now,

Substituting values,

We get,

$^5N_5 = \frac{5!}{(5-5)!}$

We get,

The total number of ways of arrangement = 5! = 5 × 4 × 3 × 2 × 1 = 120,

So,

There are 120 ways to arrange horses for riders.

Hence we can say that there are 120 ways in which 5 riders and 5 horses can be arranged.

Learn more about arrangements here

brainly.com/question/15032503

#SPJ4

7 0

1 year ago

Whats the transformation,domain and range of -x+3

Katen [24]

i’m not sure sorry about that

4 0

2 years ago