The student council at Greenwood High School is making T-shirts to sell for a fundraiser, at a price of $19 apiece. The costs, meanwhile, are $8 per shirt, plus a setup fee of $88. Selling a certain number of shirts will allow the student council to cover their costs. What will the costs be? How many shirts must be sold?

Step-by-step explanation:

je........porte

tu.......portes

nous.......portons

vous........portez

ils......portent

Answer:

rotation

Step-by-step explanation:

Answer:

Reflection

Step-by-step explanation:

I think its reflection because if you reflect them across the y axis then the x axis I believe they will be in those positions.

Answer:

i belive its A

Step-by-step explanation: hope this helps :)

Answer:

B. 4

Step-by-step explanation:

4 times 5 is 20 and 20 divided by 5 is 4

Answer:

Inequality: 30+12x < 100

x ≈ 5 months

Step-by-step explanation:

To solve, first create the inequality. In this problem, 30 dollars is the flat fee for joining the gym and thus a constant. Additionally, 12 is the monthly payment and thus the coefficient for a variable because it changes with the number of months. Finally, 100 is the most that can be spent (the max), so it should be on the other side of the inequality and the sign should be less than or equal to 100.

This makes the inequality: 30+12x < 100

To find the number of months solve the inequality for x.

So, the inequality equals x<5.8. However, the question asks for the number of full months. And, no more than 100 dollars can be spent. So, while you would normally round up. In this case, you must round down to 5 months.

Answer:

3000

Step-by-step explanation:

[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:

no

yes

yes

no

no

yes

Step-by-step explanation:

a prime number cant be devided by any thing other than itself or one

Answer:

Step-by-step explanation:

Is 2 is 3 is 11 is 18 is 20 is 27

Answer:

9 = [tex]\frac{n}{54}[/tex] or n ÷ 54

Step-by-step explanation:

nine equals the quotient of a number and 54

       /\

9 = the quotient of a number and 54

             /\

9 =       will be division of a number and 54

                                /\

                    9 = [tex]\frac{n}{54}[/tex] or n ÷ 54

Have a nice day!

    I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

Answer:

Option A: (It's just that the y-intercept and slope are in different spots.)

Line in (y=mx+b) form is y = 2x + 4

Slooe is 2

y-intercept is 4

Step-by-step explanation:

Slope-intercept form is written as y=mx+b, where m is slope and b is y-intercept.

By looking at the graph, we can already tell that the y-intercept is positive 4, since the line crosses the y-axis at that point.

Now use rise/run to find slope. (Remember that rise is movement up or down, and run is movement left or right.) With this in mind, the slope is 4/2, and if we simplify it, it's 2/1 (or basically just 2). Because we go upwards 4 units and to the right 2 units.

Hope this helps :)

Answer:

Step-by-step explanation:

Our even numbers from 1-9 are:

The largest possible number using the even numbers once is 8642.

Hoped this helped

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].

This should be right , if any doubt please comment

Answer:

|-4| < |-5|

Step-by-step explanation:

because if modules is given sub sign will be deducate

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

[tex]\fbox \colorbox{black}{ \colorbox{white}{C} \:  \:  \:   \:  \:  \:  \: \: \colorbox{white}{  = }  \:  \:  \:  \:  \:   + \colorbox{white}{6/5}}[/tex]

[tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]

Let's find the value of c ~

Answer:

c= -6

Step-by-step explanation:

2c -6 = 3c

we move all to the left

2c -6 (3c)= 0

add everything and the variables

-1c - 6 = 0

move all terms containing c to the left everything else to the right

-1c = 6

c=6/-1

c=-6

Answer:

27s

Step-by-step explanation:

d = distance to rendezvous point in miles

t = time of arrival at 10mph, i.e. 3 mins later than appointed time

10 mph = 10 miles per hour = ¹/₆ miles per min

12 mph = ¹/₅ miles per min

t = d/(¹/₆)

t = 6d

5 min = ¹/₁₂ hr

t - ¹/₁₂ = d/(¹/₅)

t = 5d + ¹/₁₂

Elimination of t:

6d = 5d + ¹/₁₂

d = ¹/₁₂ mi

t = 6(¹/₁₂)

t = ¹/₂ min

Time left to appointed time = ¹/₂ - ³/₆₀

= ¹⁰/₂₀ - ¹/₂₀

= ⁹/₂₀.min → 27 seconds

Answer:

27 min

Step-by-step explanation:

rsm

Answer:

The answer 19\24 is not reasonable because ​5\6​ is closer to 1 than to ​1\2​, and ​3\8​ is close to ​1\2​, making the difference close to 1\2. ​

Step-by-step explanation:

Answer:

1/2 1 0

Step-by-step explanation:

y=3x+6.

y=3x-5

I did it with a parallel line

Answer:

There would be 12

Step-by-step explanation:

Answer:

y=1/2x

Step-by-step explanation:

Answer:

10

9

8

7

6

5

4

3

2

1

0 1 2 3 4 5 6 7 8 9 10

Step-by-step explanation:

So the arrow is pointing at 10 and  5

Answer 10 5

Not 5

its 10 5

So the answer is 10 5

Answer:

$14.40 is the discount price.

Step-by-step explanation:

0.2 x 18 = 3.6

18 - 3.6 = 14.4

Step-by-step explanation:

f(x) = 2x² + 5√(x - 2)

Domain

√(x - 2) ≥ 0

x - 2 ≥ 0

x ≥ 2

x is greater than or equal to 2

Answer: Greater than

Step-by-step explanation:

Answer:

18, 19, 20

Step-by-step explanation:

(n) + (n + 1) + (n + 2) = 57

3n + 3 = 57

3n = 57 - 3 = 54

n = 54/3 = 18

three numbers are 18, 18+1, 18+2

18, 19, 20

Answer:

18,19,20

Step-by-step explanation:

x + (x+1) + (x+2) = 57

x + x + 1 + x + 2 = 57

Combine like terms

3x + 3 = 57

subtract 3 from both sides

3x = 54

divide both sides by 3

x = 18

The answer is 18, 19, 20

Since we know the value of x, we can look at it like this:

x + (x+1) + (x+2) --- > 18 + (18+ 1) + (18+2) --> 18 + 19 + 20

Answer:

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

Answer:

Yes, This is a repeating decimal

No, This is a decimal that neither terminates or repeats

No, This is a decimal that neither terminates or repeats

Yes, This is a terminating decimal

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]