If |x| = 6 then x = -6 or

The dice are illustrations of patterns

The fifth die that is different from the other four dice is die D

From the question, we have the following highlights

Hence, the fifth die that is different from the other four dice is die D

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The value in dollars of a stack of dimes that is 10 cm high is $10.

A dollar is further divided into some majorly used parts as:

A cent = 100th part of a dollar = $1/100 = $0.01

A nickle= 20th part of a dollar = $1/20 = $0.05

A dime = 10th part of a dollar = $1/10 = $0.10

A quarter = 4th part of a dollar = $1/4 = $0.25

If each dime is 1 mm then 10 dimes will be 1 cm high;

1 dime / 1 mm = x dimes / 10 mm Cross multiply

1 dime * 10 mm = x * 1 mm

10 dimes = x The mm

Since a dime is about 1 mm thick (10mm equals 1 cm), 10 cm high would be 100 dimes;

10 dimes = $1

$1 times 10 = $10

Hence, The value in dollars of a stack of dimes that is 10 cm high is $10.

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15 + 1.5x = 255

and 160 laps

15 ft = 15*12 = 180 inches

180/4 = 45

 so the scale is 1/45

The epicenter is located at (-5, 1) on the coordinate plane.

To find the coordinates of the epicenter, we need to find the average of the coordinates of the three receiving stations. The x-coordinate of the epicenter is (3 - 13 - 5)/3 = -5, and the y-coordinate is (3 - 5 + 3)/3 = 1. Therefore, the epicenter is located at (-5, 1) on the coordinate plane.

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The coordinates of the epicenter of earthquake are (-1, 0).

The epicenter of the earthquake is the point that is equidistant from stations X, Y, and Z. This is essentially the solution to a system of three equations representing the distances from the epicenter to each of the three stations.

The equations are derived from the distance formula, [tex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex] ,

where [tex](x_1,y_1)[/tex] and [tex](x_2, y_2)[/tex] are the coordinates of the two points and d is the distance between them.

For station X at (3,3), the distance to the epicenter (x,y) is 5 units. So, the equation is:

[tex](x - 3)^2 + (y - 3)^2 = 5^2 = 25 \text{ ....(1)}[/tex]

For station Y at (-13,-5), the distance to the epicenter is 13 units. So, the equation is:

[tex](x + 13)^2 + (y + 5)^2 = 13^2 = 169\text{ ....(2)}[/tex]

For station Z at (-5,3), the distance to the epicenter is 5 units. So, the equation is:

[tex](x + 5)^2 + (y - 3)^2 = 5^2=25 \text{ ....(3)}[/tex]

Solving this system of equations will give the coordinates of the epicenter.

Subtract equation (1) from equation (2), we get:

[tex](x^2+26x+169+y^2+10y+25)-(x^2-6x+9+y^2-6y+9) = 144\\\\\text{Simplifying, we get:}\\32x+16y+176=144\\32x+16y = -32\\\text{Divide by 16 on both sides of the equation, we get:}\\2x + y = -2[/tex]

Subtract equation (1) from equation (3), we get:

[tex](x+5)^2+(y-3)^2-(x-3)^2-(y-3)^2 = 0\\x^2+10x+25-(x^2-6x+9)=0\\16x + 16 = 0\\\text{Dividing by 1 on both sides, we get}\\x + 1 =0\\\text{We get:}\\x = -1[/tex]

Putting x = -1 in the equation 2x + y = -2, we get:

2(-1) + y = -2

We get,

y = 0

So, the coordinates of the epicenter of earthquake are (-1, 0).

The given problem consists of solving two inequalities. The first one, r4<−5, is invalid since the 4th power of any real number cannot be a negative number. For the second inequality, −2r−7 ≤ 3, the solution is r ≥ -5.

The subject of this question is Mathematics, specifically relating to solving inequalities. To answer your question, let's solve the two given inequalities independently. Starting with the first one, r4<−5, this inequality is invalid as the 4th power of any real number (r) cannot be less than -5. Moving on to the second inequality, −2r−7 ≤ 3, we first add 7 to both sides to isolate the term with the variable, giving us -2r ≤ 10. Dividing both sides by -2 (remember to flip the inequality sign when multiplying or dividing by a negative number) gives us r ≥ -5. Therefore, the solution to the given inequalities is r ≥ -5.

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To find the number of train cars and sections of wooden track in a full train set, we can set up a system of equations. The equations are 0.4x + 0.2y = 5.2 and 0.6x + 0.1y = 4.2.

To find the number of train cars and number of sections of wooden track in a full train set, we can set up a system of equations based on the given information.

Let's use x to represent the number of train cars and y to represent the number of sections of wooden track.

The system of equations that can be used to find the number of train cars and number of sections of wooden track in a full train set is:

0.4x + 0.2y = 5.2

0.6x + 0.1y = 4.2

Assuming that the cardboard forms a regular rectangle, therefore the area of the cardboard is simply taken to be the product of length and width. Therefore:

A = l w

A = (9 inches) * (4 inches)

A = 36 square inches = 36 inches^2

Answer:

[tex](x+1)[/tex] and [tex](x-4)[/tex].

Step-by-step explanation:

We have been a trinomial [tex]x^2-3x-4[/tex]. We are asked to find the factors of our given trinomial.

We will use split the middle term to solve our given problem. We need to two number whose sum is [tex]-3[/tex] and whose product is [tex]-4[/tex]. We know such two numbers are [tex]-4\text{ and }1[/tex].

Split the middle term:

[tex]x^2-4x+x-4[/tex]

Make two groups:

[tex](x^2-4x)+(x-4)[/tex]

Factor out GCF of each group:

[tex]x(x-4)+1(x-4)[/tex]

[tex](x-4)(x+1)[/tex]

Therefore, factors of our given trinomial are [tex](x+1)[/tex] and [tex](x-4)[/tex].

After 7 months her savings will be 7117.64.

Step-by-step explanation:

Given :

Rate is r = 5% = 0.05

Compounding interval, n = 12

Solution :

Now,

[tex]\rm \dfrac{r}{n} = \dfarc{0.05}{12}[/tex]

[tex]\rm \dfrac{r}{n} = 0.004167[/tex]

Now first deposit duration is 7 months. Therefore, its value is  [tex]\rm a_1 = 1000\times (1.004167)^7[/tex]

Now second deposit duration is 6 months. Therefore, its value is

[tex]\rm a_2 = 1000\times (1.004167)^6[/tex]

Now third deposit duration is 5 months. Therefore, its value is

[tex]\rm a_3 = 1000\times (1.004167)^5[/tex]

and so on.

The values are in geometric sequence where

[tex]\rm a = 1000\times (1.004167)[/tex]

[tex]\rm r = 1.004167[/tex]

We know that the sum of the geometric sequence is given by,

[tex]\rm S_n = \dfrac{a (r^n-1)}{(r-1)}[/tex]    ------ (1)

Now put the values of a and r in equation (1),

[tex]\rm S_7=\dfrac{1000\times(1.004167)((1.004167)^7-1)}{(1.004167-1)}[/tex]

[tex]\rm S_7 = 7117.64[/tex]

After 7 months her savings will be 7117.64.

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Answer:

Option B is correct

Slope of the lines that passes through the given points is, [tex]\frac{-6}{5}[/tex]

Step-by-step explanation:

Given: The points (2 , 6) and (7 ,0).

Slope of a line identify the direction of a line. To find the slope, you divide the difference of the y-coordinates of two points on a line by the difference of the x- coordinates of those same two  points.

For any two points [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] the slope(m) of a line is given by;

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

Now, substitute the point (2 ,6 ) and ( 7, 0) in the slope formula we have;

[tex]m = \frac{y_2 -y_1}{x_2 -x_1}=\frac{0 - 6}{7-2}[/tex] = [tex]\frac{-6}{5}[/tex]

therefore, the slope of the line that passes through the pair of points is;   [tex]\frac{-6}{5}[/tex]

The partial derivatives fx(30,5) and ft(30,5) can be calculated using the given information, and the tangent plane approximation can be used to estimate f(33,4).

The chemical reaction time is represented by the function f(t,x), where t is the temperature in degrees Celsius and x is the quantity of a catalyst present in grams. We are given that when the temperature is 30 degrees Celsius and there are 5 grams of catalyst, the reaction takes 50 minutes. Additionally, increasing the temperature by 3 degrees reduces the time taken by 5 minutes, and increasing the amount of catalyst by 2 grams decreases the time taken by 3 minutes.

To find the partial derivative fx(30,5), we need to find the rate of change of the reaction time with respect to the quantity of catalyst, while holding the temperature constant. Using the given information, we can calculate:

fx(30,5) = (f(30,5+2)-f(30,5))/2 = (-3)/2 = -1.5

To find the partial derivative ft(30,5), we need to find the rate of change of the reaction time with respect to the temperature, while holding the quantity of catalyst constant. Using the given information, we can calculate:

ft(30,5) = (f(30+3,5)-f(30,5))/3 = (-5)/3 = -1.67

Using the tangent plane approximation, we can estimate f(33,4) by calculating the change in reaction time by adjusting the temperature to 33 degrees Celsius and the quantity of catalyst to 4 grams.

f(33,4) = f(30,5) + ft(30,5) * (33-30) + fx(30,5) * (4-5)

f(33, 4) = 50 + (-1.67) * 3 + (-1.5) * -1 = 50 - 5.01 + 1.5 = 46.49

[tex]\( f_t(30,5) = -\frac{5}{3} \)[/tex], [tex]\( f_x(30,5) = -1.5 \)[/tex], [tex]\( f(33,4) \approx 46.5 \)[/tex] minutes.

ft (30, 5) =_[tex]\(-\frac{5}{3} \)[/tex]_ fx (30,5) =_-1.5_ f (33,4)=__46.5 minutes__

To find the partial derivatives [tex]\( f_t(30,5) \)[/tex] and [tex]\( f_x(30,5) \)[/tex], we'll use the given information about how changes in [tex]\( t \) and \( x \)[/tex] affect the time [tex]\( f(t,x) \)[/tex].

Given:

- [tex]\( f(30,5) = 50 \)[/tex] minutes

- Increasing [tex]\( t \)[/tex] by 3 degrees Celsius reduces [tex]\( f \)[/tex] by 5 minutes.

- Increasing [tex]\( x \)[/tex] by 2 grams decreases [tex]\( f \)[/tex] by 3 minutes.

1. Partial derivative with respect to [tex]\( t \) at \( (30,5) \)[/tex]:

  - Change in [tex]\( t \): \( \Delta t = 3 \)[/tex] degrees Celsius

  - Change in [tex]\( f \): \( \Delta f = -5 \)[/tex] minutes

  - Using the definition of partial derivative:

    [tex]\[ f_t(30,5) = \frac{\Delta f}{\Delta t} = \frac{-5}{3} = -\frac{5}{3} \text{ minutes/degree Celsius} \][/tex]

2. Partial derivative with respect to [tex]\( x \) at \( (30,5) \)[/tex]:

  - Change in [tex]\( x \): \( \Delta x = 2 \)[/tex] grams

  - Change in [tex]\( f \): \( \Delta f = -3 \)[/tex] minutes

  - Using the definition of partial derivative:

    [tex]\[ f_x(30,5) = \frac{\Delta f}{\Delta x} = \frac{-3}{2} = -1.5 \text{ minutes/gram} \][/tex]

Now, to find [tex]\( f(33,4) \)[/tex] using the tangent plane approximation, we'll use the partial derivatives we found and the point [tex]\( (30,5) \)[/tex] as a reference:

The tangent plane approximation formula is:

[tex]\[ f(t,x) \approx f(30,5) + f_t(30,5)(t-30) + f_x(30,5)(x-5) \][/tex]

Substituting the values:

[tex]\[ f(33,4) \approx 50 + \left(-\frac{5}{3}\right)(33-30) + (-1.5)(4-5) \][/tex]

[tex]\[ f(33,4) \approx 50 - 5 + 1.5 \][/tex]

[tex]\[ f(33,4) \approx 46.5 \text{ minutes} \][/tex]

So, [tex]\( f(33,4) \approx 46.5 \)[/tex] minutes.

Thus:

- [tex]\( f_t(30,5) = -\frac{5}{3} \)[/tex] minutes/degree Celsius

- [tex]\( f_x(30,5) = -1.5 \)[/tex] minutes/gram

- [tex]\( f(33,4) \approx 46.5 \)[/tex] minutes

The correct question is:

The number of minutes taken for a chemical reaction if f(t,x). It depends on the temperature t degrees Celsius, and the quantity, x grams, of a catalyst present. When the temperature is 30 degrees Celsius and there are 5 grams of catalyst, the reaction takes 50 minutes. Increasing the temperature by 3 degrees reduces the time taken by 5 minutes. Increasing the amount of catalyst by 2 grams decreases the time taken by 3 minutes. Use this information to find the partial derivatives fx(30,5) and ft(30,5). Use the tangent plane approximation to find f(33,4). ft (30, 5) =__________ fx (30,5) =__________ f (33,4)=__________

divide 400 by 24

400 / 24 = 16.666

 so they would need to buy 17 crates