Show Work! A warehouse worker shipped 289 boxes of notebooks. Each box contained 36 notebooks. What was the total number of notebooks shipped?

(A) 2, 601
(B) 9, 404
(C) 10, 404
(D) 10, 413

Answer:

869/50

Step-by-step explanation:

17.38

= 1738/100

= 869/50

Answer:

500000

Step-by-step explanation:

The given differential equation is

x ² (y' - x ²) + 3xy = cos(x)

Expanding and rearranging terms, we get

x ² y' + 3xy = cos(x) + x ⁴

Multiply both sides by x, which is motivated by the fact that (x ³)' = 3x ².

x ³ y' + 3x ²y = x cos(x) + x ⁵

The left side is the derivative of a product:

(x ³y)' = x cos(x) + x ⁵

Integrate both sides with respect to x :

∫ (x ³y)' dx = ∫ (x cos(x) + x ⁵) dx

x ³y = cos(x) + x sin(x) + 1/6 x ⁶ + C

Solve for y. Since x > 0, we can safely divide both sides by x ³.

y = cos(x)/x ³ + sin(x)/x ² + 1/6 x ³ + C/x³

Answer:

1st row 56 pennies

2nd row 36 pennies

3rd row 14 dimes

4th row 4 quarters

5th row 5 nickels

Step-by-step explanation:

1st row $1.85 + 56 cents = $2.41

2nd row $2.05 + 36 cents = $2.41

3rd row is $1.01 + $1.40 = $2.41

4th row $1.41 + $1.00 = 2.41

5th row $2.16 + 25 cents = $2.41

Answer:

4 cm by 5 cm (4 x 5)

Step-by-step explanation:

The area of a rectangle with length [tex]l[/tex] and width [tex]w[/tex] is given by [tex]A=lw[/tex]. Since the length of the rectangle is 7 less than 3 times its width, we can write the length as [tex]3w-7[/tex]. Therefore, substitute [tex]l=3w-7[/tex] into [tex]A=lw[/tex]:

[tex]A=lw,\\20=(3w-7)w[/tex]

Distribute:

[tex]20=3w^2-7w[/tex]

Subtract 20 from both sides:

[tex]3w^2-7w-20=0[/tex]

Factor:

[tex](w-4)(3w+5)=0,\\\begin{cases}w-4=0, w=\boxed{4},\\3w+5=0, 3w=-5, w=\boxed{-\frac{5}{3}}\end{cases}[/tex]

Since [tex]w=-\frac{5}{3}[/tex] is extraneous (our dimensions cannot be negative), our answer is [tex]w=4[/tex]. Thus, the length must be [tex]20=4l, l=\frac{20}{4}=\boxed{5}[/tex] and the dimension of the rectangle are 4 cm by 5 cm (4 x 5).

Answer:

[tex]m\angle H = 30^o[/tex]

Step-by-step explanation:

Given

See attachment

Required

Find [tex]m\angle H[/tex]

To calculate [tex]m\angle H[/tex], we make use of:

[tex]\cos(\theta) = \frac{Adjacent}{Hypotenuse}[/tex]

So, we have:

[tex]\cos(H) = \frac{GH}{HI}[/tex]

This gives:

[tex]\cos(H) = \frac{10\sqrt3}{20}[/tex]

[tex]\cos(H) = \frac{\sqrt3}{2}[/tex]

Take arccos of both sides

[tex]m\angle H = cos^{-1}(\frac{\sqrt3}{2})[/tex]

[tex]m\angle H = 30^o[/tex]

Answer:

E(XY)=1/18

Step-by-step explanation:

x y P(x,y) xy*P(X,Y)

0 0 4/9 0

0 1 2/9 0

1 0 2/9 0

1 1 1/18 1/18

2 0 1/36 0

0 2 1/36 0

1 1/18

from above:

E(XY)=1/18

Answer:

4th option

Step-by-step explanation:

6/sin(65) = 5/sin(x)

or, 6×sin(x) = 5×sin(65)

or, sin(x) = 5×sin(65)/6

or, x = arcsin(5×sin(65)/6)

Answer:

m║n

Step-by-step explanation:

If two lines 'line m' and 'line n' are perpendicular to the 'line t', both the lines 'm' and 'n' will be parallel to each other.

If m ⊥ l and n ⊥ l, then m║n.

There are 18,679,680 different hands of 5 cards where only one is a queen.

There are 52 cards. We want to make hands of 5, where only one of the cards is a queen.

There are 4 queens, so the other 48 cards are not queens.

Let's say that the first card must be a queen (order does not matter), there are 4 options to choose from.

The second card must not be a queen, so here we have 48 options.

The third card, again, must be different than a queen, so here we have 47 options (because one was already chosen).

For the fourth and fifth cards, the reasoning is similar, the number of options are 46 and 45 respectively.

The total number of combinations is given by the product between the numbers of options, we get:

C = 4*48*47*46*45 = 18,679,680

There are 18,679,680 different hands of 5 cards where only one is a queen.

If you want to learn more about combinations:

https://brainly.com/question/11732255

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Answer:

h< 0 and k > 0

Step-by-step explanation:

Quadrant 2 is when x< 0 and y > 0

For the center to be in quadrant 2

h< 0 and k > 0

Answer:

y = 3x^2-6x+3    one real solution  

y = -x^2 - 4x + 7     two real solution

y = -2x^2+9x-11       two complex solutions

Step-by-step explanation:

b^2-4ac = 0   1 repeated real solution

b^2-4ac > 0   2 distinct real solutions

b^2-4ac < 0   2 complex solutions

The quadratic functions have the following solutions:

y = 3x²-6x+3 has two real solutions.

y = -x² - 4x + 7 has one real solution.

y = -2x²+9x-11 has one complex solution.

The given quadratic functions are y = 3x²-6x+3, y = -x² - 4x + 7 and y = -2x²+9x-11.

The discriminant of a quadratic equation ax² + bx + c = 0 is in terms of its coefficients a, b, and c. i.e., Δ OR D = b² − 4ac.

Now, with the function y = 3x²-6x+3, we get

b² − 4ac=(-6)²-4×3×3=36-36=0

Since b=0 it has two real solutions.

Now, with the function y = -x² - 4x + 7, we get

b² − 4ac= (-4)²-4×(-1)×7=16+28=44

Since b>0 it has one real solutions.

Now, with the function y = -2x²+9x-11, we get

b² − 4ac= (9)²-4×(-2)×(-11)=81-88=-7

Since b<0 it has one complex solution.

Therefore, the quadratic functions have the following solutions:

y = 3x²-6x+3 has two real solutions.

y = -x² - 4x + 7 has one real solution.

y = -2x²+9x-11 has one complex solution.

To learn more about the quadratic function solutions visit:

https://brainly.com/question/1687230.

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Answer:

Number of planes on the runway or waiting to take off is approximately 2

Step-by-step explanation:

Given the data in the question;

On average there are 21 planes taking off per hour

rate of flow = frequency of take off = 21 planes / hr

= 21 planes per 60 minutes

= 0.35 planes/min

Now, we get the throughput time

throughput time = total time for take off =  waiting time on runway + run time on runway

= (4 minutes and 21 seconds) + 27 seconds

= 4.35 minutes + 0.45 minutes

= 4.8 minutes

Now, using Little's law;

Number of planes on the runway or waiting to take off will be;

N = Rate of flow × throughput time

we substitute

N = ( 0.35 planes/min ) × 4.8 min

N = 1.68 planes ≈ 2 planes

Therefore, Number of planes on the runway or waiting to take off is approximately 2

Answer:

Option A, 80π

Step-by-step explanation:

4²×5π

= 80π

Answer:

4x + 3 = 59

x = 14

Step-by-step explanation:

The vertical angles theorem states that when two lines intersect, the angles opposite each other are congruent. One can apply this here by stating the following:

4x + 3 = 59

Solve for (x), use inverse oeprations:

4x + 3= 59

4x = 56

x = 14

Answer:

Relationship : Vertical angle

Step-by-step explanation:

(4x + 3) = 59

4x = 59 - 3

4x = 56

x = 56/4

x = 14

Answer:

[tex]\pink{\bigstar}[/tex] [tex]\small\underline{\boxed{\bf\green{ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} }}}[/tex]

Answer:

qualitative

Step-by-step explanation:

bcos the question is in quality format

Answer:

we are armysss!!!!\

hiiiiiiiiii

yoooooooo

heyyyyyy

brainlist meeee!

Answer:

Y =4X  -3

Step-by-step explanation:

x1 y1  x2 y2

1 1  -2 -11

(Y2-Y1) (-11)-(1)=   -12  ΔY -12

(X2-X1) (-2)-(1)=    -3  ΔX -3

slope= 4          

B= -3          

Y =4X  -3    

Answer:

y=4x-3

Step-by-step explanation:

Hi there!

We are given the points (1,1) and (-2, -11) and we want to write the equation of the line in slop-intercept form

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

So let's find the slope of the line

The formula for the slope calculated from two points is [tex]\frac{y_2-y_1}{x_2-x_1}[/tex], where [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] are points

We have everything we need to calculate the slope, let's just label the points to avoid confusion

[tex]x_1=1\\y_1=1\\x_2=-2\\y_2=-11[/tex]

Now substitute those values into the formula

m=[tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

m=[tex]\frac{-11-1}{-2-1}[/tex]

Subtract

m=[tex]\frac{-12}{-3}[/tex]

Divide

m=4

So the slope of the line is 4

Here is the equation of the line so far:

y=4x+b

We need to find b

As the equation passes through both (1,1) and (-2, -11), we can plug either one of them into the equation to solve for b

Taking (1,1) will give us this:

1=4(1)+b

Multiply

1=4+b

Subtract 4 from both sides

-3=b

Substitute -3 as b into the equation

y=4x-3

Hope this helps!

Answer:

[tex]P(O\ or\ A) = 0.85[/tex]

Step-by-step explanation:

Given

See attachment

Required

[tex]P(O\ or\ A)[/tex]

From the question, we understand that she can only get blood from O or A groups. So, the probability is represented as:

[tex]P(O\ or\ A)[/tex]

This is calculated as:

[tex]P(O\ or\ A) = P(O) + P(A)[/tex]

Using the American row i.e. the blood must come from an American.

We have:

[tex]P(O) = 0.45[/tex]

[tex]P(A) = 0.40[/tex]

So, we have:

[tex]P(O\ or\ A) = 0.45 + 0.40[/tex]

[tex]P(O\ or\ A) = 0.85[/tex]

Answer:

20 + 10 = blank × (4 + 2)

30 = blank × 6

blank = 30 ÷ 6 = 5

Answer:

C 9i

D -9i

Step-by-step explanation:

sqrt(-81)

sqrt(81) sqrt(-1)

we know that sqrt(-1) = i

±9i

Answer:

68% of the time, John's grades will be between 55 and 85, while for Jane, 68% of the time, her grades will be between 65 and 75. They have the same mean grade, however, due to the lower standard deviation, Jane is more consistent, while John has the higher upside.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

John:

Mean of 70, standard deviation of 15.

70 - 15 = 55

70 + 15 = 85

68% of the time, John's grades will be between 55 and 85.

Jane:

Mean of 70, standard deviation of 5.

70 - 5 = 65

70 + 6 = 75.

68% of the time, Jane's grades will be between 65 and 75.

Describe the two students in terms of consistency of their grades and give reason.

68% of the time, John's grades will be between 55 and 85, while for Jane, 68% of the time, her grades will be between 65 and 75. They have the same mean grade, however, due to the lower standard deviation, Jane is more consistent, while John has the higher upside.

Answer:

Step-by-step explanation:

d = 3/2

a₂₀ = a₁+19d

35/2 = a₁ + 19×3/2

a₁ = 35/2 - 19×3/2 = -11

a₁₅ = a₁+14d = -11 + 14×3/2 = 10

Answer:

1 = - 4 - 14 √3

2 = 9 - 11 √3

Step-by-step explanation:

Question 1

(-4√3 + 2)(√3 + 4)

Apply FOIL method

= (-4√3) √3 + (-4√3) . 4 + 2 √3 + 2 . 4

Apply minus-plus rules: + (-a) = -a

= -4 √3 √3 - 4 . 4 √3 + 2 √3 + 2 . 4

Simplify

= - 4 - 14 √3

Question 2

(-3 + √3)(1 + 4 √3)

Apply FOIL method

= (-3) . 1 + (-3) . 4 √3 + √3 . 1 + √3 . 4 √3

Apply minus-plus rules: + (-a) = -a

= -3 . 1 - 3 . 4 √3 + 1 . √3 + 4 √3 √3

Simplify

= 9 - 11 √3

Answer:

The answer is below

Step-by-step explanation:

The equation of a straight line is given by:

y = mx + b;

where y, x are variables, m is the rate of change and b is the y intercept.

a) d is on the x axis and T is on the y axis, the rate of change is gotten using the points (0, 75) and (300, 37.5). Hence the rate of change (m) is:

[tex]m=\frac{y_2-y_1}{x_2-x_1} =\frac{37.5-75}{300-0} =-0.125[/tex]

b) The rate of change means that the quantity of gas in tank decreases by 0.125 for every km traveled.

c) The graph was plotted using geogebra online graphing tool.

d) Using the points (0, 75) and (300, 37.5), the equation of the line is:

[tex]T-T_1=\frac{T_2-T_1}{d_2-d_1}(x-x_1) \\\\T-75=\frac{37.5-75}{300-0}(d-0)\\\\T=-0.125d+75[/tex]

The tank is empty when T = 0, hence:

0 = -0.125d + 75

0.125d = 75

d = 600 km

The tank is empty at 600 km

Answer:

-2

Step-by-step explanation:

GIVEN :-

x = 5 and y = 12

TO FIND :-

2x - y

SOLUTION :-

placing the values of x and y

(2 × 5) - 12

10 - 12

-2

Answer: