As an inventory manager, you must decide on the order quantity for an item. Its annual demand is 679 units. Ordering costs are $7 each time an order is placed, and the holding cost is 10% of the unit cost. Your supplier provided the following price schedule. Quantity Price per Unit 1 - 100 $5.65 101 - 350 $4.95 351 or more $4.55 What ordering-quantity policy do you recommend

Answer: 8

Step-by-step explanation:

As corresponding sides of similar triangles are proportional,

[tex]\frac{RX}{RB}=\frac{XE}{BO}\\\\\frac{6}{15}=\frac{a}{20}\\\\a=20\left(\frac{6}{15} \right)=\boxed{8}[/tex]

i ) 3(2x - 3) = 4(2x + 4)

[tex]6x - 9 = 8x +1 6 \\ \\ 6x - 8x = 16 + 9 \\ \\ - 2x = 25 \\ \\ x = - \frac{25}{2} .[/tex]

ii )2(x + 2) + 5(x + 5) = 4(x-8) + 2(x - 2)

[tex]2x + 4 + 5x + 25 = 4x - 32 + 2x - 4 \\ \\ 7x + 29 = 6x - 36 \\ \\ 7x - 6x = - 36 - 29 \\ \\ x = - 65.[/tex]

2. 3x /4 - 7 /4 = 5x + 12

[tex] \frac{3x}{4} - \frac{7}{4} = 5x + 12 \\ \\ \frac{3x}{4} - 5x = 12 + \frac{7}{4} \\ \\ \frac{3x - 20x}{4} = \frac{48 + 7}{4} \\ \\ \frac{ - 17x}{4} = \frac{55}{4} [/tex]

cancel 4 from both sides

[tex] - 17x = 55 \\ \\ x = \frac{ - 17}{55} .[/tex]

3. x +7 - 8x/3 = 17/6 - 5x/8

[tex]x + 7 - \frac{8x}{3} = \frac{17}{6} - \frac{5x}{8} [/tex]

put x = 4

[tex]4 + 7 - \frac{ 8\times 4}{3} = \frac{17}{6} - \frac{5 \times 4}{8} \\ \\ 11 - \frac{32}{3} = \frac{17}{6} - \frac{20}{8} \\ \\ \frac{33 - 32}{3} = \frac{68 - 60}{24} \\ \\ \frac{1}{3} = \frac{8}{24} \\ \\ \frac{1}{3} = \frac{1}{3} .[/tex]

L.H.S. = R.H.S.

4. Let 3 consecutive integers : x , x+1 , x+2

As per the question : 2x + 3(x+1) + 4(x+2) = 56

9x + 11 = 56

Transpose 11 to RHS

9x = 45

Divide both sides by 9

x = 5 ,

x + 1 = 5 + 1 = 6

x + 2 = 5 + 2 = 7

Numbers are 5 , 6 and 7 .

5. Let the age of Jane’s younger sister is x.

Age of Jane will be x + 6

As per the question, after 10 years the sum of their ages will be 50.

After 10 years,

age of Jane = x + 16

age of her younger sister = x + 10

Therefore,

x + 16 + x +10 = 50

2x + 26 = 50

2x = 24

x = 12

Thus, the present age of Jane’s younger sister is 12 years.

And of Jane’s is 12 + 6 = 18 years.

6. Let the breadth of rectangular swimming pool be x metres.

Then, its length = (2x + 2) metres.

We know that the Perimeter of a rectangle

= 2(Length + Breadth)

= 2(2x + 2 + x) = 6x + 4

According to the given condition, we have

6x + 4 = 154

6x = 154 − 4

6x = 150

[Dividing both sides by 6]

x = 25

Thus, breadth = 25 m and

length = 25 × 2 + 2 = 52 m .

7. Let the digit in ten's .

Place be x and the digit in the one's place be y.

x − y = 5 ⟶ (i)

Two digit number = 10x + y

Two digit after reversing the digits = 10y + x

According to question ,

10x + y + 10y + x = 99

11x + 11y = 99

x + y = 9 ⟶ (ii)

On adding (i) and (ii),

2x = 14

x = 7

Putting the value of x in equation (i)

x − y = 5

y = 2

Number = 10x + y

=72 .

8. Let the speed of the steamer in still water be x km/h.

Then , the speed downstream = (x+2) km/h

and the speed upstream = (x−2) km/h

Given , distance covered in 4 hours downstream = distance covered in 5 hours upstream

4(x + 2) = 5(x − 2)

4x + 8 = 5x − 10

4x − 5x = − 10 − 8

[Transposing 5x to LHS and 8 to RHS]

−x = −18 or x = 18 km/h .

9. Let the number of notes of different denomination be x.

∴ Numbers of Rs 100 notes , 50 notes and Rs 10 notes will be 2x , 3x and 5x respectively.

Amount of 100 Rs notes ⇒Rs 100 × 2x = 200x

Amount of 50 Rs notes ⇒Rs 50 × 3x = 150x

Amount of 10 Rs notes ⇒Rs 10 × 5x = 50x

As per the question

Total amount = 400000

200x + 150x + 50x = 400000

400x = 400000 , x = 1000

No of Rs 100 notes = 2x = 2 × 1000 = 2000

No of Rs 50 notes = 3x = 3 × 1000 = 3000

No of Rs 10 notes = 5x = 5 × 1000 = 5000 .

10. Let the age of the man be x

Then age of his son becomes (x − 24)

2 years later from now,

Age of man will be = x + 2

and age of his son will be = (x − 24 + 2) = x − 22

According to question,

2(x − 22) = x + 2

i.e., 2x − 44 = x + 2

i.e., x = 46

Therefore ,

Present age of man = 46 years

And present age of his son = 46 − 24 = 22 years .

Answer:

There is no line plot

Step-by-step explanation:

Answer:

3x* 2x^2 +3x*5 -2 * 2x^2  -2 *5

Step-by-step explanation:

(3x-2)(2x^2+5)

We can FOIL to find the expansion

First 3x* 2x^2 = 6x^3

Outer 3x*5 = 15x

Inner -2 * 2x^2 = -4x^2

Last -2 *5 = -10

Add all the terms together

3x* 2x^2 +3x*5 -2 * 2x^2  -2 *5

Answer:

3x • 2x^2 + 3x • 5 + (–2) • 2x^2 + (–2) • 5

Step-by-step explanation:

other answer was right

Answer:  A B C

Step-by-step explanation:

A whole number is like 1,500 and a nonwhole number is like 33.2 or like 55/100.  

And have a good day.

Answer:

It doesn't say anything about "Length" and "Width".

Step-by-step explanation:

Brainlest . Just press the .

Answer: We can take 2 common we could also write this as 2 times X plus y.

Step-by-step explanation:

Answer:

y - 5 = -2(x + 3)

Step-by-step explanation:

When you write an equation in point-slope form, you only need two things: a point and a slope.

Given:

point: (-3, 5)

slope (m): -2

The standard point-slope equation is

y - y₁ = m(x - x₁)

Plug in what you know.

y - (5) = -2(x - (-3))

Simplify.

y - 5 = -2(x + 3)

This is your equation.

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Answer:

C = 2√(πA) units

Step-by-step explanation:

Your answer is correct. The two "bases" of the trapezoid are the line segments (0, 0)-to-(y, 0) and (0, x)-to(z, x), with lengths y and z, respectively. Then the average of the bases is [tex]\frac{y+z}2[/tex]. (We know they're the bases because the line segments are parallel.)

We multiply this by the height, which is given by the length of the line segment (0, 0)-to-(0,x), or x.

Hence the area is

[tex]\dfrac{x(y+z)}2[/tex]

(which is identical to what you have).

Answer:

m=0 , b=41

Step-by-step explanation:

Answer:

I believe you forgot to add a picture

[tex]\huge\underline\red{Answer -}[/tex]

[tex]\bold{8 \frac{2}{5} }\: \\\\ \implies \: \frac{5 × 8 + 2}{5} \: \\ \\ \implies \: \frac{ 40 + 2}{5} \\\\ \implies \frac{42}{5} \\\\ \rightarrow \:\boxed {reciprocal \: = \frac{5}{42} } \\ \\[/tex]

hope helpful ~

[tex]\huge\text{Hey there!}[/tex]

[tex]\mathsf{8 \dfrac{2}{5}}\\\\\mathsf{= \dfrac{8\times5+2}{5}}\\\\\mathsf{= \dfrac{40 + 2}{5}}\\\\\mathsf{= \dfrac{42}{5}}\\\\\mathsf{=\dfrac{5}{42}}\\\\\huge\text{Therefore, the answer should be: \boxed{\mathsf{\dfrac{5}{42}}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

43

Step-by-step explanation:

because if you add it up thats how you get that

Answer:

f'(x) = 7x^4 (5 ln(x) + 1) + x^2

Explanation:

[tex]\sf Let \ f(x)=7x^5 ln(x) + \dfrac{1}{3} x^3[/tex]

Properties used while differentiating:

[tex]i) \quad \sf \dfrac{d}{dx} (ln(x)) = \dfrac{1}{x}[/tex]

[tex]ii) \quad \sf \dfrac{d}{dx} (x^n) = n(x)^{n-1}[/tex]

[tex]iii) \quad \sf \dfrac{d}{dx}(uv)=u \dfrac{dv}{dx}+v \dfrac{du}{dx}[/tex]

Begin solving:

f'(x) =

[tex]\Longrightarrow \sf \dfrac{d}{dx}\:\left(7x^5\:ln\left(x\right)\:+\:\dfrac{1}{3}\:\:x^3\right)[/tex]

[tex]\Longrightarrow \sf \dfrac{d}{dx}\:\left(7x^5\:ln\left(x\right)\right) + \dfrac{d}{dx}\left(\dfrac{1}{3}x^3\right)[/tex]

[tex]\Longrightarrow \sf \dfrac{d}{dx}(7x^5)\:\ \times ln\left(x\right) + \dfrac{d}{dx}(ln(x)) \times \ 7x^5 + \left(3\:\times \dfrac{1}{3}x^{3-1}\right)[/tex]

[tex]\Longrightarrow \sf 5(7x^{5-1})\:\times\ ln(x) + \dfrac{1}{x} \:\times\ 7x^5 \ + x^2[/tex]

[tex]\Longrightarrow \sf 35x^{4} ln(x) + \ 7x^4 \ + x^2[/tex]

[tex]\Longrightarrow \sf 7x^4(5 ln(x) + 1) \ + x^2[/tex]

Answer:

a= 30

b=60

c=105

Step-by-step explanation:

A forms a vertical angle with another angle that measures 30 degrees, therefore,

a= 30 degrees.

Since a is 30 degrees and a and b are in the same triangle with a 90-degree angle and all interior angles in a triangle equal 180 degrees.

Then, 30+90+b=180

120+b=180

b=60

Angle c and an angle that equals 75 degrees are supplementary. meaning that they add up to 180 degrees.

180-75=c

105=c

Using proportions, it is found that the coordinates of the treasure as given as follows: (7.6, 8.8).

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

Researching the problem on the internet, the coordinates are as follows:

The treasure is buried between a rock and a tree in a 5:9 ratio, hence the expression is:

[tex]Ts - R = \frac{5}{14}(Tr - R)[/tex]

Hence, for the x-coordinate:

[tex]x - 3 = \frac{5}{14}(16 - 3)[/tex]

x - 3 = 5 x 13/14

x = 7.6.

For the y-coordinate:

[tex]y - 2 = \frac{5}{14}(21 - 2)[/tex]

y - 2 = 5 x 19/14

y = 8.8.

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Answer: the answer is b

Step-by-step explanation:

First lets simplify to slope intercept (y=mx+b)

y=[tex](2x-7)/-3[/tex]

y=-2/3x+7/3

Point 1: (0,7/3) as when x is 0, y=7/3.
Point 2: lets plug x in as 3 to cross cancel.

[tex]\frac{-2}{3} *\frac{3}{1} = \frac{-2}{1}[/tex]

So (3,-2).

Lets do point 3: plug in -3.

[tex]\frac{-2}{3} *\frac{-3}{1} = \frac{2}{1} = 2[/tex]

So (-3,2).

Your three points are (0,2.33 repeat) or (0,7/3), (3,-2) and (-3,2)

Hopefully this helps you out, heart if it was helpful and please do not hesitate to ask any questions. I have also attached an image of the line

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Answer:

Charles Babbage

Step-by-step explanation:

Answer: Hope this may help you

Charles Babbage, I think

Step-by-step explanation:

Charles Babbage is sometimes referred to as the father of computers.

Answer:

5,208,469

Step-by-step explanation:

5,000,000 + 200,000+ 8,000 + 400+60+9

5,000,000

  200,000

    00,000

       8,000

          400

            60

              9

5,208,469

If m VB = 60°  and m BS = 30°, then m ∠3 =15°.

A tangent is a line passing by touching the perimeter of the circle, perpendicular to the line joining the center and touch point.

Given, PB tangents PV, PU secants

If m VB = 60°  and m BS = 30°, then m ∠3 =

The measurement of the external angles is the semi difference of the arcs it consists.

∠3 = 1/2 (60 - 30)

∠3 = 15°

Thus, If m VB = 60°  and m BS = 30°, then m ∠3 =15°.

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The correct statement compares the means of the data in the dot plots is option C; the mode is greater for Mrs. Ramos’s class.

Median represents the middle value of the given data when arranged in a particular order.

Mode is that number that appears most frequently in the given set of data.

Given;

The Number of pets in Mr. McIntyre’s class ;

0,0,0,0,0,1, 1, 1, 1, 1, 4, 4, 4, 4, 1,

Median = 1

Mode = 0 and 1

The Number of pets in Mrs. Ramos’s Class;

0, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 3, 3, 3

Median = 3

Mode = 4 and 5

Hence, C; The mode is greater for Mrs. Ramos’s class.

The remaining part of the question is

"Which statement correctly compares the measures of center of the data in the dot plots?

The median is greater for Mr. McIntyre’s class.

The median is the same for both classes.

The mode is greater for Mrs. Ramos’s class.

The mode is the same for both classes."

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Answer:

$16,400

Step-by-step Explanation:

Step 1: We multiply $10,000 by 16%, getting $1,600.

Step 2: We then multiply $1,600 by 4 (which is given). That gives us $6,400.

Step 3: Add that to $10,000. We get $16,400 as the answer!

Answer:

Simple interest = P  x  r  x  t

                         = 10,000  x  16/100  x   4

                         =  $6400

Final amount = P (1 +rt)

                       = 10,000 (1 + 16/100  x  4)

                       = $16400

(or you can just add the simple interest to the principle amount to get the final amount)

Answer: 10

Step-by-step explanation:

[tex]\frac{5}{x+5} (2x-5(2))\\\frac{5}{x+5} (2x-10)\\\frac{10x-50}{x+5}\\\frac{10(x+5)}{x+5} \\\frac{10}{1} \\10[/tex]

The difference of two squares expression is (d) 25a^2-36b^6

The difference of two squares is represented as:

x^2 - y^2

Where x and y are perfect square expressions.

From the list of options, we have:

25a^2-36b^6

The terms of the above expression are perfect squares

i.e.

25a^2 = (5a)^2

36b^6 = (6b^3)^2

Hence, the difference of two squares expression is (d) 25a^2-36b^6

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Answer:

[tex]4\sqrt{26}[/tex]

Step-by-step explanation:

The distance formula is [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex].

Thus we can plug the numbers in:

[tex]\sqrt{(-2-(-6))^2+(10-(-10))^2}[/tex]

[tex]\sqrt{(4)^2+(20)^2}[/tex]

[tex]\sqrt{16+400}[/tex]

[tex]\sqrt{416}[/tex]

[tex]4\sqrt{26}[/tex]

Answer:

So, the answer is 20

Step-by-step explanation:

[tex]f(x) = \frac{1}{2} x + 10 = = = > \\ f(x) - 10 = \frac{1}{2} x = = = > \\ 2f(x) - 20 = x = = = > \\ {f}^{ - 1} (y) = 2y - 20 = = = > {f}^{ - 1} (x) = 2x - 20[/tex]

The missing value of the inverse function is 20.

Option D is the correct answer.

A function has an input and an output.

A function can be one-to-one or onto-one.

It simply indicated the relationships between the input and the output.

Example:

f(x) = 2x + 1

f(1) = 2 + 1 = 3

f(2) = 2 x 2 + 1 = 4 + 1 = 5

The outputs of the functions are 3 and 5

The inputs of the function are 1 and 2.

We have,

To find the inverse of a function, we usually replace f(x) with y, then solve for x in terms of y, and finally replace y with the inverse function notation h(x).

So let's do that with the given function:

f(x) = (1/2)x + 10

y = (1/2)x + 10

2y = x + 20 (multiply both sides by 2 to isolate x)

x = 2y - 20 (subtract 20 from both sides)

Now we can replace x with h(x) and y with x to get the inverse function:

h(x) = 2x - 20

Thus,

The missing value is 20.

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Answer:

yes

Step-by-step explanation:

because bank

Answer:

80 students are female

Step-by-step explanation:

21 are male meaning 19 are female

9 are female in French

12 are female in Spanish

total = 40

2x + 40 = 120

2x = 80

x = 40

females'  = 80

total = 80

Step-by-step explanation:

look at the attachment above

Answer:

[tex]\textsf{1)} \quad -\dfrac{1}{16}e^{-4x}\left(4x+1\right)+\text{C}[/tex]

[tex]\textsf{2)} \quad - \cos x+\dfrac{2}{3} \cos^3 x - \dfrac{1}{5} \cos^5 x +\text{C}[/tex]

Step-by-step explanation:

Question 1

[tex]\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Integration of $e^{ax}$} \\\\$\displaystyle \int e^{ax}\:\text{d}x=\dfrac{1}{a}e^{ax}+\text{C}$\\\\for $a\neq 0$\\\end{minipage}}[/tex]

Given integral:

  [tex]\displaystyle \int xe^{-4x}\:\text{d}x[/tex]

Using Integration by parts:

[tex]\textsf{Let }\:u=x \implies \dfrac{\text{d}u}{\text{d}x}=1[/tex]

[tex]\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=e^{-4x} \implies v=-\dfrac{1}{4}e^{-4x}[/tex]

Therefore:

[tex]\begin{aligned}\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x & =uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x\\\\\implies \displaystyle \int xe^{-4x}\:\text{d}x & =-\dfrac{1}{4}xe^{-4x}-\int -\dfrac{1}{4}e^{-4x}\: \text{d}x\\\\& =-\dfrac{1}{4}xe^{-4x}+\int \dfrac{1}{4}e^{-4x}\: \text{d}x\\\\& =-\dfrac{1}{4}xe^{-4x}-\dfrac{1}{16}e^{-4x}+\text{C}\\\\& =-\dfrac{1}{16}e^{-4x}\left(4x+1\right)+\text{C}\end{aligned}[/tex]

Question 2

[tex]\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\\ \end{minipage}}[/tex]    

[tex]\boxed{\begin{minipage}{5 cm}\underline{Integrating a constant}\\\\$\displaystyle \int n\:\text{d}x=nx+\text{C}$\\(where $n$ is any constant value)\end{minipage}}[/tex]

Rewrite the given integral:

[tex]\begin{aligned}\displaystyle \int \sin^5 x \: \text{d}x & =\int (\sin x)^4 \cdot \sin x \: \text{d}x\\& =\int (\sin^2 x)^2 \cdot \sin x \: \text{d}x\end{aligned}[/tex]

Use the trig identity  [tex]\sin^2x+\cos^2x \equiv 1[/tex]  to rewrite  [tex]\sin^2x[/tex] :

[tex]\implies \displaystyle \int \sin^5 x \: \text{d}x = \int (1-\cos^2 x)^2 \cdot \sin x \: \text{d}x[/tex]

Integration by substitution

[tex]\textsf{Let }\:u=\cos x \implies \dfrac{\text{d}u}{\text{d}x}=-\sin x \implies \text{d}x=-\dfrac{1}{\sin x}\: \text{d}u[/tex]

Therefore:

[tex]\begin{aligned}\implies \displaystyle \int \sin^5 x \: \text{d}x & = \int (1-u^2)^2 \cdot \sin x \cdot -\dfrac{1}{\sin x}\: \text{d}u\\& = \int -(1-u^2)^2 \: \text{d}u\\ & =\int -1+2u^2-u^4 \: \text{d}u\\& =-u+\dfrac{2}{3}u^3-\dfrac{1}{5}u^5+\text{C}\end{aligned}[/tex]

Finally, substitute  [tex]u = \cos x[/tex]  back in:

[tex]\implies \displaystyle \int \sin^5 x \: \text{d}x=- \cos x+\dfrac{2}{3} \cos^3 x - \dfrac{1}{5} \cos^5 x +\text{C}[/tex]