add or subtract the polynomials
(8x3+3x-2)+(9x2-7+3x-2x3)

Answer:

ok so.u grit da he on 40/ rock cause u got a andriodnh on gf

Answer:

9 hours worked after school and 4 hours worked on saturday

Step-by-step explanation:

x = amount of hours worked after school

y = amount of hours worked on saturday

x + y = 13 or x = 13 - y

and

3x + 4y = 43

plug:

3 · (13 - y) + 4y = 43

39 - 3y + 4y = 43

y = 4

plug:

x = 13 - y

x = 13 - 4 = 9

9 hours worked after school and 4 hours worked on saturday

Answer:

95m

Step-by-step explanation:

312 - 61 - 61 = 190

190/2 = 95

Answer:

What is the common ratio of the sequence 3, 21, 147, …?  

7

Which formula can be used to find the nth term of the sequence 3, 21, 147, …?

c)

Use the given formula to find the indicated terms of the sequence.

a4 = 1029

a5 = 7203

Step-by-step explanation:

Answers for all three questions<3

[tex](4a - 5b)^{2} \\ by \: \: \: using \: \: \: (x - y)^{2} = {x}^{2} - 2xy + {y}^{2} \\ = {(4a)}^{2} - 2(4a)(5b) + {(5b)}^{2} \\ = {16a}^{2} - 40ab + 25 {b}^{2} [/tex]

[tex] {16a}^{2} - 40ab + {25b}^{2} [/tex]

Hope it helps.

Do comment if you have any query.

Step-by-step explanation:

cost per day

= $1500 / week

= $1500 / 7 days

= $1500 ÷ 7 / 7 ÷ 7

≈ $214,29 / day

Answer:

[tex]\frac{\sqrt{70} }{5}[/tex]

Step-by-step explanation:

= f( g(2) )

= 3(2x + 1) + 1

= 6x + 3 + 1

= 6x + 4

= 6(2) + 4

= 12 + 4

= 16

f(g(2)) = 16

9514 1404 393

Answer:

  63

Step-by-step explanation:

The number of seats in a row will give an arithmetic sequence:

  6, 9, 12, 15, ...

The first term is 6; the common difference is 3. The general term is ...

  an = a1 +d(n -1) . . . . . . n-th term of sequence with first term a1, difference d

The 20th term of the sequence is ...

  a20 = 6 +3(20 -1) = 6 +57 = 63

There would be 63 seats on the 20th row.

Answer: 9

Step-by-step explanation:

36 ÷ 4 = 9

Answer:

z

Step-by-step explanation:

hqvqhqvw karrar ras wallah a part

Answer:

x=g-c

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]2\sqrt{8}+3\sqrt{50}=2\sqrt{2*2*2} + 3\sqrt{5*5*2}\\\\= 2*2\sqrt{2}+3*5\sqrt{2}\\\\=4\sqrt{2}+15\sqrt{2}\\\\= 19\sqrt{2}[/tex]

[tex]3\sqrt{12}+4\sqrt{27}=3\sqrt{2*2*3}+4\sqrt{3*3*3}\\\\=3*2\sqrt{3}+4*3\sqrt{3}\\\\=6\sqrt{3}+12\sqrt{3}\\\\=(6+12)\sqrt{3}\\\\= 18\sqrt{3}[/tex]

Answer:

Step-by-step explanation:

To find the y-intercept, set x=0 and evaluate the function.

  f(0) = -3/(0 -4) = 3/4

The y-intercept is (0, 3/4).

__

To find the x-intercept(s), set f(x) = 0 and solve for x.

  0 = -3/(x^2 -4)

  0 = -3 . . . . . . . . . . multiply by (x^2 -4), x ≠ ±2

There are no values of x that will make this true. There are no x-intercepts.

_____

Additional comments

In general, you find the x-intercepts of a rational function by finding the zeros of the numerator. Here, the numerator is -3, so cannot ever be zero.

I find a graphing calculator to be a useful tool for showing where to look for x-and y-intercepts. The attached graph shows y=0 (the x-axis) is a horizontal asymptote, so there are no x-intercepts.

Answer: 9 triangles

Step-by-step explanation:

Answer:

|-4| < |-5|

Step-by-step explanation:

because if modules is given sub sign will be deducate

Answer: The second bubble thing is correct you have picked the wrong one.

Answer:

Option D)  [tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex]

Step-by-step explanation:

Given the equation, T = 15 + c × n, where:

T = represents the total monthly gym cost

c =  represents the additional charge for a group class, and

n = represents the number of group classes attended

In order to determine which equation can be used to find the number of group classes a customer attended, if there are given values for c and T, we must isolate the variable, n  algebraically.

The first step is to subtract 15 from both sides:

T = 15 + c × n

T - 15 = 15 - 15 + c × n

T - 15 = c × n

Next, divide both sides by c to isolate n :

[tex]\huge\mathsf{\frac{({T\:-\:15})}{c}\:=\:\frac{{c\:\times\:n}}{c}}[/tex]

[tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex]

Therefore, the correct answer is Option D) [tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex].

Answer: i solved on my channel

Step-by-step explanation: https://youtu.be/rTgj1HxmUbg

Answer:

20

Step-by-step explanation:

20*20 = 400

OR

[tex]20 {}^{2} = 400[/tex]

Answer: i think it is y=−56x−4

Step-by-step explanation:

The equation in the slope intercept form which passes through the points ( 0, -4 ) and ( 6 , 9 ) is y = (-5 / 6)x - 4.

The correct answer is Option C.

Given data:

To find the equation of a line in slope-intercept form (y = mx + b) that passes through the points (0, -4) and (6, -9), we need to determine the slope (m) and the y-intercept (b).

First, calculate the slope (m):

m = (change in y) / (change in x)

m = (-9 - (-4)) / (6 - 0)

m = (-9 + 4) / 6

m = -5 / 6

Now that we have the slope, we can use one of the given points (let's use (0, -4)) to solve for the y-intercept (b):

-4 = (-5 / 6) * 0 + b

-4 = b

So, the y-intercept (b) is -4.

Now, we can write the equation of the line in slope-intercept form:

y = (-5 / 6)x - 4

Hence, the equation of the line is y = (-5 / 6)x - 4.

To learn more about equation of line, refer:

https://brainly.com/question/14200719

#SPJ3

The complete question is attached below:

Which equation, in slope-intercept form, matches the equation shown?

a line that goes through the points (0, -4) and (6, -9)

A) y = ( 4/7 )x - 4

B) y = ( 5/6 )x + 1

C) y = ( -5/6 )x - 4

D) y = ( -4/7 )x + 1

Answer:

yes

7164 is divisible bt 6

Answer:

Yes

Step-by-step explanation:

Any number will be divisible by 6 if they be divisible by 2 & 3

- we know 7164 is divisible by 2

- also any number which sum of their digits are divisible by 3, the number will be divisible by 3

in this case (7164) 7+1+6+4=18 & 18 is divisible by 3 so 7164 is divisible by 3 as well.

so 7164 is divisible by 6

Answer:

.......................

Answer:

a) 2 * 3 = 6.

b) 123 is odd but contains an even digit (2).

Step-by-step explanation:

Answer:

x = 49

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are in proportion, that is

[tex]\frac{DN}{KJ}[/tex] = [tex]\frac{DG}{KM}[/tex] , substitute values

[tex]\frac{5}{35}[/tex] = [tex]\frac{8}{x+7}[/tex] ( cross- multiply )

5(x + 7) = 280 ( divide both sides by 5 )

x + 7 = 56 ( subtract 7 from both sides )

x = 49

Step-by-step explanation:

[tex] = {( \frac{27}{8} )}^{ \frac{1}{3} } \times ( \frac{243}{32} )^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = { ({ (\frac{3}{2} )}^{3}) }^{ \frac{1}{3} } \times {( {( \frac{3}{2}) }^{5} )}^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{3 \times \frac{1}{3} } \times {( \frac{3}{2} )}^{5 \times \frac{1}{5} } \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = \frac{3}{2} \times \frac{3}{2} \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{1 + 1 + 2} [/tex]

[tex] = {( \frac{3}{2} )}^{4} \: or \: \frac{81}{16} [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{27}{8} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{243}{32} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

We can write as :

27 = 3 × 3 × 3 = 3³

8 = 2 × 2 × 2 = 2³

243 = 3 × 3 × 3 × 3 × 3 = 3⁵

32 = 2 × 2 × 2 ×2 × 2 = 2⁵

[tex]\sf{\longmapsto{\bigg( \dfrac{3 \times 3 \times 3}{2 \times 2 \times 2} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{{(3)}^{3}}{{(2)}^{3}} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{({3}^{5})}{{(2)}^{5}} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now, we can write as :

(3³/2³) = (3/2)³

(3⁵/2⁵) = (3/2)⁵

[tex]\sf{\longmapsto{\left\{\bigg(\frac{3}{2} \bigg)^{3} \right\}^{\frac{1}{3}} \times \Bigg[\left\{\bigg(\frac{3}{2} \bigg)^{5} \right\}^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now using law of exponent :

[tex]{\sf{({a}^{m})^{n} = {a}^{mn}}}[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{3 \times \frac{1}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{5 \times \frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{\frac{3}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{\frac{5}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times\Bigg[\bigg(\frac{3}{2} \bigg)^{1} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \times \dfrac{3}{2} \bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3 \times 3}{2 \times 2}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)\times \Bigg[\bigg(\frac{3}{2} \bigg)\times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3}{2} \times \dfrac{9}{4} \: \: \Bigg]}}\\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3 \times 9}{2 \times 4} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg(\dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{27}{8} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3}{2} \times \dfrac{27}{8}}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3 \times 27}{2 \times 8}}} \\[/tex]

[tex] \sf{\longmapsto{\dfrac{81}{16}}\: ≈ \:5.0625\:\red{Ans.}} \\[/tex]

Answer:

30

Step-by-step explanation:

3 snow cones per 0.5 pounds of ice

0.5 pounds of ice x 10 = 5 pounds of ice

3 snow cones X 10 = 30 snow cones.