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Charra [1.4K]
3 years ago
12

3xy + 5x + 2 + 3y +x + 4

Mathematics
2 answers:
Leto [7]3 years ago
6 0
3xy+6x+3y+6 is the answe
Aleks04 [339]3 years ago
5 0

Answer:

= 3xy + ( 5x + x) + 3y + ( 4 +2 )

= 3xy + 6x + 3y + 6

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Choose Yes or No to tell if the fraction 3/5 will make each equation true.
Vinvika [58]
Yes
no
no
yes
i hope it helped.
4 0
3 years ago
Read 2 more answers
Can someone fr help me with this like right now i’ll appreciate it sm
rjkz [21]

Step-by-step explanation:

Since

\frac{108}{72}  =  \frac{3}{2}

\frac{72}{48}  =  \frac{3}{2}

\frac{48}{32}  =  \frac{3}{2}

The function has a constant factor so we have a exponential function.

The equation of a exponential function is

y = a(b) {}^{x}

Where a is the initial value and b is the growth factor

The growth factor is 3/2,

The initial value in 32

y =  32 ( \frac{3}{2} ) {}^{x}

Plug in 4 for x.

You will get 162

Plug in 5 for x you will get

243.

2013- 162

2014- 243

8 0
2 years ago
Which equation is represented by the graph below? HELP PLEASE
bonufazy [111]

Answer: The answer is b

Step-by-step explanation:

7 0
3 years ago
Suppose that prices of a certain model of a new home are normally distributed with a mean of $150,000. Use the 68-95-99.7 rule t
levacccp [35]

Answer:

68% of buyers paid between $147,700 and $152,300.

Step-by-step explanation:

We are given that prices of a certain model of a new home are normally distributed with a mean of $150,000.

Use the 68-95-99.7 rule to find the percentage of buyers who paid between $147,700 and $152,300 if the standard deviation is $2300.

<u><em>Let X = prices of a certain model of a new home</em></u>

SO, X ~ Normal(\mu=150,000 ,\sigma=2,300)

The z score probability distribution for normal distribution is given by;

                             Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean price = $150,000

            \sigma = standard deviation = $2,300

<u>Now, according to 68-95-99.7 rule;</u>

Around 68% of the values in a normal distribution lies between \mu-\sigma and \mu-\sigma.

Around 95% of the values occur between \mu-2\sigma and \mu+2\sigma .

Around 99.7% of the values occur between \mu-3\sigma and \mu+3\sigma.

So, firstly we will find the z scores for both the values given;

         Z  =  \frac{X-\mu}{\sigma}  =  \frac{147,700-150,000}{2,300}  = -1

         Z  =  \frac{X-\mu}{\sigma}  =  \frac{152,300-150,000}{2,300}  = 1

This indicates that we are in the category of between \mu-\sigma and \mu-\sigma.

SO, this represents that percentage of buyers who paid between $147,700 and $152,300 is 68%.

7 0
4 years ago
If -3 is a lower bound for the real zeros of f(x), then -2 is also a lower bound.
vladimir1956 [14]

The statement is false, because -2 is larger than -3.

<h3>Is the statement true?</h3>

We know that -3 is a lower bound for the real zeros of f(x).

So, if we define k as the smallest real zero of f(x), we have:

-3 < k.

Then we can take, for example, k = -2.5

That is an allowed value for the real zero.

In that case, -2 is not a lower bound, because is false that:

-2 < k = -2.5

Then if -3 is a lower bound for the real zeros of f(x), we can't assume that -2 is also a lower bound (because -2 is larger than -3).

If you want to learn more about lower bounds:

brainly.com/question/15991714

#SPJ1

4 0
2 years ago
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