Third-degree, with zeros of −3, −1, and 2, and passes through the point (1,8).

Answer:

Hi,

[tex]u_{20}=30[/tex]

Step-by-step explanation:

[tex]u_n=u_1+(n-1)*r, \ r= commun\ ratio.\\u_4=u_1+3*r=6\\u_6=u_1+5r\\(u_6)^2=u_1^2+10ru_1+25r^2\\s_n=u_1+u_1+r+u_1+2r+...+u_1+10r=11*u_1+55r\\s_{11}=(u_6)^2+18\ \\\Longrightarrow\ u_1^2+10ru_1+25r^2+18=11u_1+55r\\\Longrightarrow\ 4r^2+2r-12=0\\\Longrightarrow\ 2r^2+r-6=0\\r=\dfrac{3}{2} \ or\ r=-2\ (excluded)\\\\u_1=6-3r=\dfrac{3}{2} \\\\\\\boxed{u_{20}=u_1+19*r=30}\\[/tex]

The value of the 20th term of the arithmetic series will be 30

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

Un  =  u₁+(n-1) x r

u₄  =  u₁  +(3 x r)=6

u₆ = u₁ +5r

u₆²=u₁²+10ru₁+25r²+18=11u₁+55r

Sn= u₁+u₁+r+u₁+2r+.............u₁+10r= (11 x u₁) +55 r

4r²+2r-12=0

2r²+r-6=0

r=3/2

u₁=6-3r=3/2

u₂₀=u₁+(19 x r)= 30

Hence the value of the 20th term of the arithmetic series will be 30

To know more about arithmetic progression follow

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Answer:

If you draw the graph of y = -x + 6  on the above graph  the solution to system will be the coordinates of the point of intersection.

the equation of the line drawn above is y = 0.5x

so you can also find the solution by solving the system

y = 0.5x

y = -x + 6

0.5x = -x + 6

1.5x = 6

x = 6/1.5 = 4

and y = 0.5*4 = 2

The answer is  (4, 2)     - the third choice.

Step-by-step explanation:

Answer:

1 : 5

Step-by-step explanation:

3 students zoo

15 students aquarium

zoo : not zoo

3: 15

Divide each side by 3

3/3 : 15/3

1 : 5

Answer:

30 quarts

Step-by-step explanation:

4 quarts = 1 gallon

7.5 gallons * 4 quarts/ 1 gallon = 30 quarts

Answer:

8

Step-by-step explanation:

By power of a point, (x-2)(x+13) = (x-3)(x+16). Solving, we find that x=11, so FG= 11-3 = 8

Answer: 2.31

Step-by-step explanation:

Answer:

25

Step-by-step explanation:

product of 18 and -3=18*-3= -54

positive difference=52-29=23

(18x-3)+56+(52-29)

= -54 + 56 + 23

=25

Answer: 318 km²

Let's find the areas of the triangles

Formula: base x height / 2

9 x 6 = 54

54/2 = 27

Multiply by 2 since there are two triangles (I know, we already got the areas from 9 x 6)

27 x 2 = 54

Now let's try to find the areas of all the rectangular sides

Formula: length x width

9 x 11 = 99

8 x 11 = 88

7 x 11 = 77

99 + 88 + 77 = 264

Add up all the areas

264 + 54 = 318

Answer:

-60 and -28

Step-by-step explanation:

-60 + 42= -18

CHECK:

-42-18=-60

-56+-28=-84

CHECK:

-84+28=-56

Hope this helps <3 Need thanks? Comment /hearthelp

Answer:

-60

-28

Step-by-step explanation:

x +42 = -18

Subtract 42 from each side

x+42-42 = -18-42

x = -60

-56 +y = -84

Add 56 to each side

y = -84+56

y = -28

Use the law of cosines, which says

189² = 215² + 123² - 2×215×123 cos(∠P)

==>   cos(∠P) = 25,633/52,890

==>   ∠P ≈ 61.06°

Answer:

1) 6а-9b=3(2a-3b)

2)8a-12b=4(2a-3b)

3) 5ab-5ac=5a(b-c)

4) 6ax+6ay=6a(x+y)

Answer:

Hello,

Step-by-step explanation:

Formula used (a+b)^2=a²+2ab+b²

[tex](12-9i)^2=3^2*(4-3i)^2\\\\=9*(16-9-24i)\\\\=9*(7-24i)\\\\=63-216i\\[/tex]

Answer:

37500

Step-by-step explanation:

pattern is multiplying by -5.

-7,500*-5 = 37,500

Have a nice day! :-)

Answer:

3√13

Step-by-step explanation:

3√13

This expression cannot be simplified because you cannot take any common factor outside the root.

Let y = √x, so the limit can be rewritten as

[tex]\displaystyle \lim_{y\to\infty}\left(\frac{\sqrt{y^6+y^2}}{y^2+3} - \frac{\sqrt{y^4+1}}{y+4}\right)[/tex]

Now,

[tex]\sqrt{y^6+y^2} = \sqrt{y^2\left(y^4+1\right)} = y\sqrt{y^4+1}[/tex]

so we can rewrite the limit further as

[tex]\displaystyle \lim_{y\to\infty}\left(\frac{y}{y^2+3} - \frac1{y+4}\right)\sqrt{y^4+1}[/tex]

Combine the rational terms:

[tex]\dfrac y{y^2+3} - \dfrac1{y+4} = \dfrac{y(y+4)-(y^2+3)}{(y^2+3)(y+4)} = \dfrac{4y-3}{(y^2+3)(y+4)}[/tex]

Then in the limit, we get

[tex]\displaystyle \lim_{y\to\infty}\frac{(4y-3)\sqrt{y^4+1}}{(y^2+3)(y+4)} = \lim_{y\to\infty}\frac{(4y^3-3y^2)\sqrt{1+\dfrac1{y^4}}}{y^3+4y^2+3y+12} \\\\ = \lim_{y\to\infty}\frac{\left(4-\dfrac3y\right)\sqrt{1+\dfrac1{y^4}}}{1+\dfrac4y+\dfrac3{y^2}+\dfrac{12}{y^3}} = \boxed{4}[/tex]

Answer:

B). 34.85°

Step-by-step explanation:

From trigonometric ratios:

[tex] \sin(C) = \frac{opposite}{hypotenuse} [/tex]

Opposite = 16

Hypotenuse = 28

[tex] \sin(C) = \frac{16}{28} \\ \\ C = { \sin}^{ - 1} ( \frac{16}{28} ) \\ \\ C = 34.85 \degree[/tex]

Answer:

19

Step-by-step explanation:

Add 7 onto 12 to find answer.

Answer:

b = 19

Step-by-step explanation:

b = 12 + 7

Simplify 12+7 to 19.

Answer:

B, E

Step-by-step explanation:

( 9 × 100 )  +  ( 2 × 10 )  +  ( 3 × 0.10 )  +  ( 5 ×0.01 )  = 920.35

A.  92.35

B.  920.350

C.  902.35

D.  92.350

E.  920.35

F. 920.035

Answer:

0.36

Step-by-step explanation:

Answer:

i don't know

( lol )

heheeeee

Answer:

12 yrs.

Step-by-step explanation:

let present age of girl=x

age of mother=x+18

after 6 yrs

age of girl=x+6

age of mother=x+18+6=x+24

according to the condition of question

(x+6)+(x+24)=54

2x+30=54

2x=54-30=24

x=24/2=12

Answer:

Sara is now 4, will be 8 in 4 years. Ralph is now 12, will be 16 in 4 years.

Answer:  [tex]\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right)[/tex]

We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.

===========================================================

Explanation:

It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.

I'm going to use these three log rules, which apply to any base.

From there, we can then say the following:

[tex]\frac{1}{2}\log_{2}\left(m\right)-3\log_{2}\left(n\right)+2\log_{2}\left(q\right)\\\\\log_{2}\left(m^{1/2}\right)-\log_{2}\left(n^3\right)+\log_{2}\left(q^2\right) \ \text{ .... use log rule 3}\\\\\log_{2}\left(\sqrt{m}\right)+\log_{2}\left(q^2\right)-\log_{2}\left(n^3\right)\\\\\log_{2}\left(\sqrt{m}*q^2\right)-\log_{2}\left(n^3\right) \ \text{ .... use log rule 1}\\\\\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right) \ \text{ .... use log rule 2}[/tex]

Answer:

the cost of the tire=341.4$

The ratio is (J/W) = 7/6. Or. J= (7/6)• W .

Second line: J= (7/6)• 24. J= 28.

Third line: J= (7/6)• 18. J= 21

Answer:

-5.82

Possible by long devison and the rule that if there is one negitive then the answer is negitive

Using the function concept, it is found that the result would not be a function, as for each month, there would be multiple possible number of days.

In a function, one value of the input can be related to only one value of the output.

In this problem, the input is the month and the output is the possible number of days in the month.

Since there is a month with either 28 or 29 days, and the other months have either 30 or 31 days, the are multiple outputs for each input, hence, the result is not a function.

You can learn more about the function concept at https://brainly.com/question/12463448

Answer:

A

Step-by-step explanation:

The area (A) of a rectangle is calculated as

A = length × breadth

   = (2x - 5)(x + 4) ← expand using FOIL

   = 2x² + 8x - 5x - 20

  = 2x² + 3x - 20 → A

Answer:

29 feet

Step-by-step explanation:

14 × 3 = 42

100 - 42 = 58

58 ÷ 2 = 29

One of the longer chains is 29 feet long.

Hope this helps.