All categories

2 years ago

Solve the quadratic equation by factoring-

Mathematics

1 answer:

shtirl [24]2 years ago

3 0

Answer:

c- (2x+3)(x+1)

Step-by-step explanation:

Let me know if you want step by step

You might be interested in

The table shows the relationship between the number of songs Alisha has purchased and the cost of her purchase. Find the rate of

Gre4nikov [31]

Answer:

whats the question

Step-by-step explanation:

4 0

2 years ago

There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=

yulyashka [42]

Answer:

S = {0,2,3,4}

P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001

Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

P(X = 0) = $\frac{12P4}{12^{4} }$

P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = $\frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }$

P(X=2) = 0.401

P(X=3) = $\frac{4C3 * 12P2}{12^{4} }$

P(x=3) = 0.025

P(X=4) = $\frac{12}{12^{4} }$

P(X=4) = 0.001

Mean, $\mu = \sum xP(x)$

$\mu = (0*0.573) + (2*0.401) + (3*0.025) + (4*0.001)\\\mu = 0.879$

Standard deviation, $SD = \sqrt{\sum x^{2} P(x) - \mu^{2}} \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2} * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}$

SD = 1.033

4 0

3 years ago

Please show your work. i’ll give brainliest!

pshichka [43]

Factor 4x2−16x+16

4x2−16x+16

=4(x−2)(x−2)

Answer:

4(x−2)(x−2)

6 0

2 years ago

Read 2 more answers

Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Travka [436]

Answer:

I cant see the picture. could you put it in the comments

Step-by-step explanation:

5 0

3 years ago

An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

USPshnik [31]

The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

4 0

2 years ago