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3 years ago

Question: (Graph photo attached)

Mathematics

1 answer:

Artist 52 [7]3 years ago

8 0

Answer:

The line plot depicted shows, number of books read by 15 students .

Books read Number of Students

0 0

1 1

2 2

3 2

4 3

5 1

6 2

7 3

8 1

9 0

Number of students who read more than 3 books

= 3 + 1 +2+3+1+0

= 10 students

Option C

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janet completed this entier column of math homework in 10 min. How many sec per question did janet spend

xxTIMURxx [149]

Step-by-step explanation:

10 minutes = 600 seconds

Seconds per question is seconds/questions

Divide 600 by the number of questions:

600/(number of questions) = ________ seconds per question

4 0

3 years ago

WILL MARK BRAINLIEST! Can someone please help! I don't understand some of these questions :(

PolarNik [594]

Answer:

Step-by-step explanation:

The interior and exterior angle of a polygon is supplementary

let interior be I

let exterior be E

I + E = 180

Since the interior angle is 8 times that of an exterior angle,

8E + E = 180 [replacing I with 8E]

9E = 180

E = 20

The exterior angle is 20 degrees

I + E = 180

I + 20 = 180

I = 160

The interior angle is 160 degrees.

The equation to find the interior angle of a polygon with 'n' number of sides is:

I = ( (n − 2) × 180 ) ⁄ n

We know the interior angle, so plug it in and solve for n:

160 = ( (n − 2) × 180 ) ⁄ n

160n = (n − 2) × 180

160n = 180n − 360

-20n = -360

n = 18

3 0

3 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

3 years ago

Can you tell de answer?

kodGreya [7K]

Yes, it’s D because you would have want to compare both fractions by making them have the same denominator. Since 60 is the least common denominator for 10 and 12 (the denominators), you would convert the fractions by multiplying the numerator and denominator of 5/10 by 6 to get 60. 5/10 would turn into 30/60. Then you would multiply 6/12 both by 5, and then it would turn into 30/60. Now that you have equal fractions, you can guess that they are the same. Therefore, 30/60=30/60! An easier way of doing it would be just simplifying them into 1/2 and comparing them + seeing that they are both 1/2. Hope this helps!

•••

Caramelatte

7 0

3 years ago

Please help me, I already answered the first 3

Alchen [17]

Answer:

4= 29 $x^{2}$ +6

all I know is number fourr T-T I tried yw

Step-by-step explanation:

5 0

3 years ago