is this correct??????

Here are your answers.

it's made by computer.

- BRAINLIEST answerer

Answer:

z = 6

Step-by-step explanation:

- 2.3z = - 13.8 ( divide both sides by - 2.3 )

z = [tex]\frac{-13.8}{-2.3}[/tex] = 6

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.

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

the answer is -2w (i have the write at least 20 words to dont worry about this)

Answer:

-2w

Step-by-step explanation:

Like terms have same variable.

34w - 4g + 4g - 36w = 34w - 36w - 4g + 4g

                                  = - 2w + 0 = -2w

Answer:

[tex]3x - 5 = 7x + 11 \\ 7x - 3x = - 11 - 5 \\ 4x = - 16 \\x = - \frac{16}{4} \\ \color{yellow} \boxed{x =-4}[/tex]

Answer:

x= -4

Step-by-step explanation:

3x-5=7x+11

3x-5+5=7x+11+5

3x=7x+11+5

3x=7x+16

3x-7x=7x+16-7x

-4=7x+16-7x

-4x=16

divided each side by -4

x= 16 over -4

divide them and you get -4.

Answer: 9 triangles

Step-by-step explanation:

Answer:

l=5+2w

Step-by-step explanation:

Since the length is twice the width plus 5, the equation would look like :

l=5+2w

A=lw

w=width

2w+5=length

42=w(2w+5)

42=2w2+5w

2w2+5w-42=0

use the quadratic equation

(-5+√25+336)/4

(-5+√361)/4

(-5+19)/4

14/4=3.5

w=3.5 m

2w+5=12 m

3.5 by 12 rectangle

Answer:

x=g-c

Step-by-step explanation:

Answer:

f=2/3

Step-by-step explanation:

Answer:z=12

Step-by-step explanation:

Answer:

135 grams of Zinc

Step-by-step explanation:

zinc:copper

3:14

3/14=x/630

x=135

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:

x = -3

Step-by-step explanation:

Answer:

x=-3

Step-by-step explanation:

You can just do +x on both sides so: 0=3+x and then -3 on both sides: -3=x.

Answer:

The first three are functions

Step-by-step explanation:

4th one fails the Vertical line test.

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:

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

Answer:

z

Step-by-step explanation:

hqvqhqvw karrar ras wallah a part

Explanation:

The ratio we want to find can be thought of in the form in:out

We list the number of "in-state" people first followed by the "out-of-state" people next. A colon separates the two values. So we have the initial ratio of 108:63

Divide both parts by the GCF 9

The ratio 108:63 fully reduces to 12:7 meaning there are 12 in-state students for every 7 out-of-state students. The order is important because 7:12 would be incorrect and imply we had 7 in-state vs 12 out-of-state.

[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.

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:

a) 2 * 3 = 6.

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

Step-by-step explanation:

Answer:

95m

Step-by-step explanation:

312 - 61 - 61 = 190

190/2 = 95

Answer:

y=[tex]-\frac{3}{4}[/tex]x+2

Step-by-step explanation:

y=sx+2;

-1=4s+2;

s=-3/4;

y=[tex]-\frac{3}{4}[/tex]x+2

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.