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3 years ago

How do I Factor 5x+15+xy+3y

2 answers:

8 0

I saw right away that 5 is common to the left 2 terms and that y is common to the right 2 terms. Thus, 5(x+3)+y(x+3) = (x+3)(5+y)

Valentin [98]3 years ago

7 0

you split the problem into two separate problems and go from there.

so, (5x+15) and (xy+3y) then you take out the common factor in both

so, positive 5 out of (5x+15) and positive y out of (xy+3y)

you'll then have 5(x+3) and y(x+3)

The x=3 is similar in both problems so you can combine them to make one and combine the two factors you took out to make another problem.

So your final answer is (x+3)(5+y)

You might be interested in

Penny had $117, which is 9 times as much money as Kari had. How much money did Kari have?

mylen [45]

Well 117/9=13 so kari has $13

8 0

4 years ago

The population of a suburb grows at a rate proportional to the population. Suppose the population doubles in size from 3000 to 6

frozen [14]

Answer:

$P(t) = 3000e^{0.1155t}$

Step-by-step explanation:

The growth of the population can be modeled by the following differential equation:

$\frac{dP}{dt} = rP$

Where r is the growth rate, P is the population, and t is the time measures in months.

I am going to solve the above differential equation with the separation of variables method.

$\frac{dP}{P} = rdt$

Integrating both sides:

$ln P = rt + P(0)$

Where P(0) is the initial condition

We need to isolate P in this equation, so we do this

$e^{ln P} = e^{rt + P(0)}$

$P(t) = P(0)e^{rt}$

The problem states that P(0)=3000, so:

$P(t) = 3000e^{rt}$

The problem wants us to find the value of r:

It states that the population doubles in size from 3000 to 6000 in a 6- month period, meaning that P(6) = 6000. So

$6000 = 3000e^{6r}$

$e^{6r} = 2$

To isolate 6r, we apply ln to both sides.

$ln (e^{6r}) = ln 2$

$6r = 0.693$

$r = \frac{0.693}{6}$

r = 0.1155

The particular solution to the differential equation with the initial condition P(0)=3000 is:

$P(t) = 3000e^{0.1155t}$

5 0

3 years ago

47÷10+86×450 I just need to figure this out I need help

Nonamiya [84]

Use PEMDAS, which stands for Parenthesis, Exponents, Multiply, Divide, Add, and Subtract.
First multiply, so 86 x 450 = 38700, and then divide 47 divided by 10, which is 4.7, and then add 4.7 plus 38700, which would give you 38704.7

3 0

3 years ago

There are many ways to measure the reading ability of children. Research designed to improve reading performance is dependent on

ELEN [110]

Answer:

A 90% confidence interval to estimate the mean DRP score in Henrico County Schools is =32±2.7280

i.e. C.I. [29.3,34.7]

Hence the results are not significant.

Step-by-step explanation:

Given:

Total no of students =44

True mean =32

Standard deviation=11

And all 4 students score also given.

To Find:

90 % confidence interval and conclusion on it.

Solution:

Here for C.I. we required mean, standard deviation and no of students.

So mean =32,S.D.=11 and n=44

Therefore , For 90 % interval Zscore is Z=1.645

C.I.=mean±Z[Standard deviation/Sqrt(n)]

=32±1.645[11/Sqrt(44)]

=32±1.645[1.6583]

=32±2.7280

i.e. C.I. [29.3,34.7]

Now Here to commit on calculated interval we should check value of Z-score

So we need to calculate the sample mean.

Sample mean =Sum of all values /Total no of values.

=[40+ 26 +39+ 14+ 42+ 18 +25+ 43+ 46 +27 +19+ 47+ 19 +26+ 35 +34+ 15+ 44 40+ 38+ 31+ 46 +52 +25+ 35 +35+ 33 +29 +34 +41 +49+ 28+ 42+ 47 +35 +48 +22 33+ 41 +51 +27 +14+ 54 +45]/44

=34.86

Z-score=(sample mean -true mean)/[standard deviation/Sqrt(n)]

=(34.86-32)/[11/Sqrt(44)]

Z=1.72465

For ,

P(Z≥1.72465)=0.08544

For 0.05 significance level the results are not significant at p<0.01.

Researcher claimed mean is not correct upto 0.05 significant level

i.e calculated mean and Sample mean are different.

Hence the results are not significant.

6 0

3 years ago

A CHICKEN COOP HOLDS A TOTAL OF 10 HENS OR 20 CHICKS. IF 2/5THS OF THE HENS IN A FULL COOP ARE REMOVED, HOW MANY NEW CHICKS COUL

LiRa [457]

Answer:

8 new chicks can be fitted in the coop.

Step-by-step explanation:

A chicken coop holds 10 hens or 20 chicks.

That means space in the coop for 10 hens = space for 20 chicks

Or space for 1 hen = space for 2 chicks

Now $\frac{2}{5}$ th of the hens were removed from the coop.

So, number of hens removed from the coop = $\frac{2}{5}\times 10$

= 4 hens

And space for 4 hens in the coop = space for the 8 chicks

Therefore, 8 new chicks can be fitted in the coop.

5 0

3 years ago