Deondra needs to order some new supplies for the restaurant where she works. The restaurant needs at least 624 glasses. There are currently 344 glasses. If each set on sale contains 20 glasses, write and solve an inequality which can be used to determine ss, the number of sets of glasses Deondra could buy for the restaurant to have enough glasses

Answer:

......

Step-by-step explanation:

Answer:

y = (7/4)x + 0

Step-by-step explanation:

yeah-ya....... right?

Answer:

It is D

Step-by-step explanation:

I am not exactly sure how to explain but i'll try.

The 2 dollars is added to her price, so it is not A.

The 25 dollars is every day, so it is mutlipled by x. So its not B.

The 2 dollars is not everyday like the 25 dollars, so its not C b/c its not everyday.

Answer:

first let's solve for x

x + (x + 30) + 2x = 180

4x + 30 = 180

-30          -30

4x = 150

/4      /4

x = 37.5

Now solve for the angle measures:

(x + 30)    =    (37.5 + 30) = 67.5

2x       =        2(37.5) = 75

x = 37.5

Answer:

X= 37.5

Step-by-step explanation:

Answer:

-3/7,0

Step-by-step explanation:

Answer:

y=−9/7x+18/7

Step-by-step explanation:

2x+7y+7x=18

9x+7y=18

Step 1: Add -9x to both sides.

9x+7y+−9x=18+−9x

7y=−9x+18

Step 2: Divide both sides by 7.

7y/7=−9x+18/7

y=−9/7x+18/7

Given :

To Find :

Solution :

[tex]\qquad { \dashrightarrow \: { \sf{2x + 7y + 7x = 18}}}[/tex]

Adding the like terms :

[tex]\qquad { \dashrightarrow \: { \sf{ 7y + 9x = 18}}}[/tex]

Transposing 9y to the other side which then becomes negative :

[tex]\qquad { \dashrightarrow \: { \sf{ 7y = 18 - 9x}}}[/tex]

Dividing both sides by 7 :

[tex]\qquad { \dashrightarrow \: { \sf{ \dfrac{7y}{7} = \dfrac{18 - 9x}{7} }}}[/tex]

[tex]\qquad { \dashrightarrow \: { \sf{ y = \dfrac{18 - 9x}{7} }}}[/tex]

⠀

Therefore, the value of y = 18 – 9x/7 .

Answer:

I'd say none, as we're missing something in this problem. Make sure you've included everything to solve this problem. Thanks.

Answer:

Step-by-step explanation:

With the amount x of portraits the daily balance is:

The total is:

Correct choices are A and C

With the amount x of portraits the daily balance is:

- $22.00

+ $15.50x

- $3.25x

15.50x - 3.25x - 22.00 ⇒ 12.25x - 22.00

Answer:

6:4:3

Step-by-step explanation:

each can be divided by 7

Answer:

6 4 3

Step-by-step explanation:

Using the normal distribution and the central limit theorem, it is found that there is a 0.0571 = 5.71% probability of selecting a sample  of 40 two-bedroom apartments and finding the mean to be at least $275 per month.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem:

The probability of selecting a sample  of 40 two-bedroom apartments and finding the mean to be at least $275 per month is the p-value of Z when X = 275, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{275 - 250}{\frac{100}{\sqrt{40}}}[/tex]

[tex]Z = 1.58[/tex]

[tex]Z = 1.58[/tex] has a p-value of 0.9429.

1 - 0.9429 = 0.0571

0.0571 = 5.71% probability of selecting a sample  of 40 two-bedroom apartments and finding the mean to be at least $275 per month.

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Answer:

720 ft^3

Step-by-step explanation:

6 * 10 *12

v = l * w * h

6 * 10 * 12 =  720

Volume is a three-dimensional scalar quantity. The volume of the self-storage centre room is 720 ft³.

A volume is a scalar number that expresses the amount of three-dimensional space enclosed by a closed surface.

The volume of a room or box that is the shape of a rectangular prism is calculated by multiplying the length, width, and height of the prism.

Volume = Length × Width × Height

Given that the self-storage centre room's are 6 feet wide, 10 feet deep, and 12 feet high. Therefore, the dimension of the room can be written as,

Length = 10 ft

Width = 6 ft

Height = 12 ft

Now, the volume of the room can be written as,

Volume of the room = 10ft × 6ft × 12ft

                                  = 720 ft³

Hence, the volume of the self-storage centre room is 720 ft³.

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square: A=s^2

triangle: A=(1/2)bh

rectangle: A=lw

parallelogram: A=bh

Answer:

y = x - 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = 1 , then

y = x + c ← is the partial equation

To find c substitute (4, 1 ) into the partial equation

1 = 4 + c ⇒ c = 1 - 4 = - 3

y = x - 3 ← equation of line

Given that

Sin θ = a/b

LHS = Sec θ + Tan θ

⇛(1/Cos θ) + (Sin θ/ Cos θ)

⇛(1+Sin θ)/Cos θ

We know that

Sin² A + Cos² A = 1

⇛Cos² A = 1-Sin² A

⇛Cos A =√(1-Sin² A)

LHS = (1+Sin θ)/√(1- Sin² θ)

⇛ LHS = {1+(a/b)}/√{1-(a/b)²}

= {(b+a)/b}/√(1-(a²/b²))

= {(b+a)/b}/√{(b²-a²)/b²}

= {(b+a)/b}/√{(b²-a²)/b}

= (b+a)/√(b²-a²)

= √{(b+a)(b+a)/(b²-a²)}

⇛ LHS = √{(b+a)(b+a)/(b+a)(b-a)}

Now, (x+y)(x-y) = x²-y²

Where ,

On cancelling (b+a) then

⇛LHS = √{(b+a)/(b-a)}

⇛RHS

⇛ LHS = RHS

Sec θ + Tan θ = √{(b+a)/(b-a)}

Hence, Proved.

Answer: If Sinθ=a/b then Secθ+Tanθ=√{(b+a)/(b-a)}.

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the solution to the system of equation from the replacement set is -5

Given the equation −6z+1=−4−7z, we are to find the value of z from the given equation:

Given

−6z+1=−4−7z

Collect the like terms;

-6z + 7z = -4 - 1

Simplify the result

z = -4 -1

z= -5

Hence the solution to the system of the equation from the replacement set is -5

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Answer:

-5

Step-by-step explanation:

I took the quiz in k12

The probability of getting an odd number on the first roll and less than 3 on the second roll is 1/6.

It is a branch of mathematics that deals with the occurrence of a random event.

The probability of getting an odd number on the first roll is 3/6

since there are three odd numbers (1, 3, 5) out of six possible outcomes (1, 2, 3, 4, 5, 6).

The probability of getting less than 3 on the second roll is 2/6, since there are two possible outcomes (1, 2) out of six.

To find the probability of both events happening, we multiply the probabilities:

P(odd on first roll and less than 3 on second roll) = P(odd on first roll) × P(less than 3 on second roll)

= (3/6)×  (2/6)

= 1/6

Therefore, the probability of getting an odd number on the first roll and less than 3 on the second roll is 1/6.

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Answer:

x = 9

Step-by-step explanation:

i took the quiz

[tex]\mathfrak{-2(x-6)=6} [/tex]

[tex]\mathfrak{-2x+12=6} [/tex]

[tex]\mathfrak{-2x=6-12} [/tex]

[tex]\mathfrak{-2x=-6} [/tex]

[tex]\boxed{\mathfrak{x=3} }[/tex]

Answer: 50/19

Step-by-step explanation:

I'm no math genius however I believe this is division.

500/190 gave me a fraction which is 50/19.

Answer:

-5+

Step-by-step explanation:

Slope of the line passing through the points (1, 4) and (3, 2) is -1.

The slope of the line is the ratio of the rise to the run, or rise divided by the run. It describes the steepness of line in the coordinate plane.

The slope intercept form of a line is y=mx+b, where m is slope and b is the y intercept.

The slope of line passing through two points (x₁, y₁) and (x₂, y₂) is

m=y₂-y₁/x₂-x₁

The line passing through the points (1, 4) and (3, 2)

Plug in the points in the slope formula.

Slope = 2-4/3-1

=-2/2

=-1

Hence, slope of the line passing through the points (1, 4) and (3, 2) is -1.

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Answer:ITS THE SECOND ONE

Step-by-step explanation:

Answer:

(a) The particle is moving to the right in the interval  [tex](0 \ , \ \displaystyle\frac{\pi}{2}) \ \cup \ (\displaystyle\frac{3\pi}{2} \ , \ 2\pi)[/tex] , to the left in the interval [tex](\displaystyle\frac{\pi}{2}\ , \ \displaystyle\frac{3\pi}{2})[/tex], and stops when t = 0, [tex]\displaystyle\frac{\pi}{2}[/tex], [tex]\displaystyle\frac{3\pi}{2}[/tex] and [tex]2\pi[/tex].

(b) The equation of the particle's displacement is [tex]\mathrm{s(t)} \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ 3[/tex]; Final position of the particle [tex]\mathrm{s(2\pi)} \ = \ 3[/tex].

(c)  The total distance traveled by the particle is 9.67 (2 d.p.)

Step-by-step explanation:

(a) The particle is moving towards the right direction when v(t) > 0 and to the left direction when v(t) < 0. It stops when v(t) = 0 (no velocity).

Situation 1: When the particle stops.

[tex]\-\hspace{1.7cm} v(t) \ = \ 0 \\ \\ 5 \ \mathrm{sin^{2}(t)} \ \mathrm{cos(t)} \ = \ 0 \\ \\ \-\hspace{0.3cm} \mathrm{sin^{2}(t) \ cos(t)} \ = \ 0 \\ \\ \mathrm{sin^{2}(t)} \ = \ 0 \ \ \ \mathrm{or} \ \ \ \mathrm{cos(t)} \ = \ 0 \\ \\ \-\hspace{0.85cm} t \ = \ 0, \ \displaystyle\frac{\pi}{2}, \ \displaystyle\frac{3\pi}{2} \ \ \mathrm{and} \ \ 2\pi[/tex].

Situation 2: When the particle moves to the right.

[tex]\-\hspace{1.67cm} v(t) \ > \ 0 \\ \\ 5 \ \mathrm{sin^2(t) \ cos(t)} \ > \ 0[/tex]

Since the term [tex]5 \ \mathrm{sin^{2}(t)}[/tex] is always positive for all value of t of the interval [tex]0 \ \leq \mathrm{t} \leq \ 2\pi[/tex], hence the determining factor is cos(t). Then, the question becomes of when is cos(t) positive? The term cos(t) is positive in the first and third quadrant or when [tex]\mathrm{t} \ \epsilon \ (0, \ \displaystyle\frac{\pi}{2}) \ \cup \ (\displaystyle\frac{3\pi}{2}, \ 2\pi)[/tex] .  

*Note that parentheses are used to demonstrate the interval of t in which cos(t) is strictly positive, implying that the endpoints of the interval are non-inclusive for the set of values for t.

Situation 3: When the particle moves to the left.

[tex]\-\hspace{1.67cm} v(t) \ < \ 0 \\ \\ 5 \ \mathrm{sin^2(t) \ cos(t)} \ < \ 0[/tex]

Similarly, the term [tex]5 \ \mathrm{sin^{2}(t)}[/tex] is always positive for all value of t of the interval [tex]0 \ \leq \mathrm{t} \leq \ 2\pi[/tex], hence the determining factor is cos(t). Then, the question becomes of when is cos(t) positive? The term cos(t) is negative in the second and third quadrant or  [tex]\mathrm{t} \ \epsilon \ (\displaystyle\frac{\pi}{2}, \ \displaystyle\frac{3\pi}{2})[/tex].

(b) The equation of the particle's displacement can be evaluated by integrating the equation of the particle's velocity.

[tex]s(t) \ = \ \displaystyle\int\ {5 \ \mathrm{sin^{2}(t) \ cos(t)}} \, dx \ \\ \\ \-\hspace{0.69cm} = \ 5 \ \displaystyle\int\ \mathrm{sin^{2}(t) \ cos(t)} \, dx[/tex]

To integrate the expression [tex]\mathrm{sin^{2}(t) \ cos(t)}[/tex], u-substitution is performed where

[tex]u \ = \ \mathrm{sin(t)} \ , \ \ du \ = \ \mathrm{cos(t)} \, dx[/tex].

[tex]s(t) \ = \ 5 \ \displaystyle\int\ \mathrm{sin^{2}(t) \ cos(t)} \, dx \\ \\ \-\hspace{0.7cm} = \ 5 \ \displaystyle\int\ \ \mathrm{sin^{2}(t)} \, du \\ \\ \-\hspace{0.7cm} = \ 5 \ \displaystyle\int\ \ u^{2} \, du \\ \\ \-\hspace{0.7cm} = \ \displaystyle\frac{5u^{3}}{3} \ + \ C \\ \\ \-\hspace{0.7cm} = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ C \\ \\ s(0) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(0)} \ + \ C \\ \\ \-\hspace{0.48cm} 3 \ = \ 0 \ + \ C \\ \\ \-\hspace{0.4cm} C \ = \ 3.[/tex]

Therefore, [tex]s(t) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(t)} \ + \ 3[/tex].

The final position of the particle is [tex]s(2\pi) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(2\pi)} \ + \ 3 \ = \ 3[/tex].

(c)

[tex]s(\displaystyle\frac{\pi}{2}) \ = \ \displaystyle\frac{5}{3} \ \mathrm{sin^{3}(\frac{\pi}{2})} \ + \ 3 \\ \\ \-\hspace{0.85cm} \ = \ \displaystyle\frac{14}{3} \qquad (\mathrm{The \ distance \ traveled \ initially \ when \ moving \ to \ the \ right})[/tex]

[tex]|s(\displaystyle\frac{3\pi}{2}) - s(\displatstyle\frac{\pi}{2})| \ = \ |\displaystyle\frac{5}{3} \ (\mathrm{sin^{3}(\frac{3\pi}{2})} \ - \ \mathrm{sin^{3}(\displaystyle\frac{\pi}{2})})| \ \\ \\ \-\hspace{2.28cm} \ = \ \displaystyle\frac{5}{3} | (-1) \ - \ 1| \\ \\ \-\hspace{2.42cm} = \displaystyle\frac{10}{3} \\ \\ (\mathrm{The \ distance \ traveled \ when \ moving \ to \ the \ left})[/tex]

[tex]|s(2\pi) - s(\displaystyle\frac{3\pi}{2})| \ = \ |\displaystyle\frac{5}{3} \ (\mathrm{sin^{3}(2\pi})} \ - \ \mathrm{sin^{3}(\displaystyle\frac{3\pi}{2})})| \ \\ \\ \-\hspace{2.28cm} \ = \ \displaystyle\frac{5}{3} | 0 \ - \ 1| \\ \\ \-\hspace{2.42cm} = \displaystyle\frac{5}{3} \\ \\ (\mathrm{The \ distance \ traveled \ finally \ when \ moving \ to \ the \ right})[/tex].

The total distance traveled by the particle in the given time interval is[tex]\displaystyle\frac{14}{3} \ + \ \displaystyle\frac{5}{3} \ + \ \displaystyle\frac{10}{3} \ = \ \displaystyle\frac{29}{3}[/tex].

Answer:

Step-by-step explanation:

X/8 = 5

X = 40

Adding 1 to the largest number (or any other of the numbers, for that matter) makes the sum 41 and the average 41/8 or 5 1/8.

The average of a given number is the sum of the given numbers divided by the number of numbers in the given set.

The resulting average of the eight numbers after adding 1 to the largest number is 5.125.

It is the average value of the set given.

It is calculated as:

Mean = Sum of all the values of the set given / Number of values in the set

Example:

2, 3, 4, 5

The average number of 1, 2, 4, and 5 is 3.

i.e 12 ÷ 4 = 3

We have,

The average of eight different numbers is 5.

Let the sum of 8 different numbers be M.

Average = M / 8

5 = M / 8

M = 8 x 5

M = 40

Now,

If 1 is added to the largest number.

This means,

40 + 1 = 41

The resulting average of the eight numbers:

= 41/8

= 5.125

Thus,

The resulting average of the eight numbers after adding 1 to the largest number is 5.125.

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Answer:

x = -4

y = -4

Step-by-step explanation:

L1     x - 5y = 16

L2  4x - 2y = -8

4*L1 - L2

4x - 20y = 64

-4x + 2y  =    8

-----------------------

      -18y = 72

          y = 72 / -18

          y = -4

x - 5y = 16

x = 16 + 5y

x = 16 + 5(-4)

x = 16 - 20

x = -4

Answer:

v=[tex]\frac{(w+v)/t}{2}[/tex]

Step-by-step explanation:

just inverse operations and you will always get your answer :)

Answer:

8, 2.

Step-by-step explanation:

2*4 is 8 so I think line them up together and 12/6 is 2

Answer:

49%.

Step-by-step explanation:

2205      100        220500

........... x ..........  =  ..................  = 49 (49%).

4500        1              4500

The theater filled up with 49% of seats with 6th grade students.

A mathematical expression is made up of terms (constants and variables) separated by mathematical operators. A mathematical equation is used to equate two expressions. Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis, observations and results of the given problem.

We have a middle school that took all of its 6th grade students on a field trip to see a symphony at a theater that has 4500 seats. The students filled 2205 of the seats in the theater.

Assume that the theater filled up [x]% of seats at the theater. Then, we can write -

[x]% = (2205/4500) x 100

[x]% = 0.49 x 100

[x]% = 49%

Therefore, the theater filled up with 49% of seats with 6th grade students.

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