Answer: The volume of the metal cube 1x1x1 m is:
V_cube = l × w × h = 1 m × 1 m × 1 m = 1 m^3
The volume of the metal block that is to be cut from the cube is:
V_block = l × w × h = 3 cm × 3 cm × 4 cm = 36 cm^3
We need to convert the volume of the block to cubic meters, as the volume of the cube is given in cubic meters. We can use the conversion factor 1 m = 100 cm to convert cubic centimeters to cubic meters.
V_block = 36 cm^3 ÷ (100 cm/m)^3 = 0.000036 m^3
The number of blocks that can be cut from the cube is:
N_blocks = V_cube ÷ V_block = 1 m^3 ÷ 0.000036 m^3 = 27,777.78
Since we can only have a whole number of blocks, the maximum number of blocks that can be cut from the cube is 27,777.
The remaining volume of the cube after cutting the metal block is:
V_remain = V_cube - V_block × N_blocks = 1 m^3 - 0.000036 m^3 × 27,777 = 0.999 m^3
The surface area of the remaining cube can be calculated as follows:
S_remain = 6 × (l × w) = 6 × (h^2) = 6 × (0.999 m)^2 = 5.994 m^2
Therefore, the complete number of blocks that can be cut from the metal cube is 27,777. The volume of each block is 36 cm^3 or 0.000036 m^3. The remaining cube has a volume of 0.999 m^3 and a surface area of 5.994 m^2.
Step-by-step explanation:
Which angles are supplementary to <4 iready
Any angle that shares a common vertex with <4 and whose measure adds up to 180 degrees is supplementary to <4.
To determine which angles are supplementary to <4, we first need to understand the concept of supplementary angles. Supplementary angles are pairs of angles whose sum is 180 degrees. In other words, if we have two angles, and their measures add up to 180 degrees, then they are supplementary to each other.
To find angles that are supplementary to <4, we need to look for other angles that add up to 180 degrees with <4. We can do this by examining the angles that intersect <4 or that share a common vertex with <4.
One way to identify such angles is to draw a diagram of the situation. Suppose we have a line with <4 marked on it. We can then draw another line that intersects the first line at <4, creating two angles on either side of <4. Let's call these angles A and B, where A is adjacent to <4 and B is adjacent to the other side of <4.
If we can find angle A or angle B, we can determine the other angle that is supplementary to <4. We know that A and B add up to 180 degrees, so if we find one of these angles, we can subtract it from 180 degrees to find the other angle.
To summarize, to find angles that are supplementary to <4, we can:
Draw a diagram of the situation, with <4 marked on a line and another line intersecting it at <4.
Identify the adjacent angles on either side of <4.
Label these angles A and B.
Determine the measure of either A or B.
Subtract the measure of the known angle from 180 degrees to find the measure of the other angle.
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Variables x and y are connected by the relationship y = Ax" where A and n are constants. Given that A = e ^ 0.5 and n = 3/8 find the value of y when x = 11 .
When x = 11, the value of y is approximately 8.276. We are given that the relationship between x and y is given by [tex]y = Ax^n[/tex]where A and n are constants.
We are also given the values of A and n:
[tex]A = e^0.5[/tex]
n = 3/8
We can substitute these values into the equation to get:
[tex]y = e^0.5 x^(3/8)[/tex]
Now, we are asked to find the value of y when x = 11. To do this, we substitute x = 11 into the equation:
[tex]y = e^0.5 x^(3/8)[/tex]
[tex]y = e^0.5 (11)^(3/8)[/tex]
We can use a calculator to evaluate this
First, we evaluate the exponent (3/8) using the exponent rules:
[tex](11)^(3/8) = (11^(1/8))^3 = 1.706[/tex]
Next, we substitute this value back into the equation:
[tex]y = e^0.5 x^(3/8)[/tex]
[tex]y = e^0.5 (11)^(3/8)[/tex]
[tex]y = e^0.5 (1.706)[/tex]
Using a calculator, we can evaluate this expression to get:
y ≈ 8.276
Therefore, when x = 11, the value of y is approximately 8.276.
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A random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph. From past information, it is known that the standard deviation of the population is 22 mph. If we are interested in determining an interval estimate for μ at 96. 6% confidence, the z value to use is
The interval estimate for the population mean, μ, is 65 ± 3.38 mph, or 61.62 to 68.38 mph.
The z-value for a 96.6% confidence level is 2.58. This value is used to determine the margin of error for a sample mean in a population. The formula for margin of error is ME = z*σ/√n, where σ is the population standard deviation, n is the sample size, and z is the z-value for the desired confidence level. In this case, ME = 2.58*22/√121 = 3.38 mph.Therefore, the interval estimate for the population mean, μ, is 65 ± 3.38 mph, or 61.62 to 68.38 mph.
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mad sounds ugh PLEASE HELP ME THIS IS URGENT I WILL GIVE BRAINLIEST sorry ik the Pic is dark my camera is broken
Answer:
B i think
Step-by-step explanation:
b
Which graph best represents y=x^+6x-1?
Answer:
ANSWER IS PROVIDED BELOW
hope this helps and good luck on your assignment
please help i will give 20 points if added demonstration on how to graph the answer
The solution is the factor 4y - 6x:. 2(2y - 3x) = 2(2y - 3x). 8 = 2• 4 =. 2(2y - 3x)
8. 2• 4
cancel the common factor: 2
= 2y - 3x. the answer is 2y - 3x
4 4
61 x 5 = 3,050 % 10, 381 X 27 = 1,143 % 9, 854 X 63 = 7,786 % 9, or 562 X 42 = 3,272, which equation is true, I don’t have the division sign so
Step-by-step explanation:
We can simplify each equation to determine which one is true:
61 x 5 = 305
10381 x 27 = 279,387
1143 ÷ 9854 x 63 = 0.00736 x 63 = 0.46368
562 x 42 = 23,604
Therefore, the only equation that is true is 562 x 42 = 23,604.
Help me! Please (You dont need to do Both questions 1 and 2!)
1. Ben's mistake in solving the area of circle is, he has taken the value of diameter at the place of radius in formula.
2. Luna's mistake is, he did the incorrect squaring of radius 4.5.
Explain about the area of circle?The formula makes use of the diameter. The units are squared at all times. When the diameter is known, the area can be calculated by first determining the radius. The radius of the circle is the area. You can use circumference to determine area.
area of circle A = πr²
where, π = 3.14 and r is the radius of circle
r = diameter /2 = 9/2 = 4.5 cm
A = πr²
A = 3.14*(4.5)²
A = 3.14*20.25
A = 63.585 cm² (correct area)
Thus,
1. Ben's mistake in solving the area of circle is, he has taken the value of diameter at the place of radius in formula.
2. Luna's mistake is, he did the incorrect squaring of radius 4.5.
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If the triangle has sides with three different lengths what are all the possible whole number lengths for the sides that meet at a right angle
Answer:
Step-by-step explanation:
A) 6 inches. ( if the equation for the area of a triangle is
80% of ? = 60
HELP MEEE
PLSS
Answer:
48
Step-by-step explanation:
first multiply 60 by 0.80 the decimal version of the percentage then you get 48.00 or 48 as your anser
Figure ABCDEFGH is a regular octagon. Segment AH is located at A(4, 5) and H(7, 5). What is the perimeter of ABCDEFGH? (1 point)
15 units
20 units
24 units
32 units
The perimeter of ABCDEFGH is 8s, with s = 3, so the perimeter of ABCDEFGH is 8x3 = 24 units.
The perimeter of a regular octagon can be calculated using the formula P = 8s, where s is the length of each side. To solve for the perimeter of ABCDEFGH, we need to first calculate the length of each side.We can use the distance formula to calculate the length of AH. The distance formula is [tex]d = √((x2-x1)^2 + (y2-y1)^2).[/tex] In this case, x1 = 4, y1 = 5, x2 = 7, and y2 = 5. Plugging these values into the formula, we get [tex]d = √((7-4)^2 + (5-5)^2) = √(3^2 + 0) = √9 = 3[/tex] units.Since each side of a regular octagon is equal, we can use the length of AH, 3 units, to calculate the perimeter of the octagon. The perimeter of ABCDEFGH is 8s, with s = 3, so the perimeter of ABCDEFGH is 8x3 = 24 units.
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you have three empty boxes and seven balls. how many ways can you distribute the balls among the three boxes so that each box contains at least one ball?
In conclusion there are 33 ways to distribute seven identical balls into three distinct boxes so that each box contains at least one ball.
How to solve?
This is a combinatorics problem known as distributing identical objects into distinct boxes with the condition that each box has at least one object.
There are three such cases: when the first box is empty, when the second box is empty, and when the third box is empty. Each of these cases can be represented by a single bar that is placed at the beginning or end of the line of stars. For example, the case where the first box is empty can be represented as:
|****|
There are three ways to place this single bar, so we need to subtract three from our previous count of 36 to get the final answer:
36 - 3 = 33
Therefore, there are 33 ways to distribute seven identical balls into three distinct boxes so that each box contains at least one ball.
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The points Q(6, 3), R(-4,-3), S(-1,-8), and T(9,-2) form rectangle QRST.
Plot the points then click the "Graph Quadrilateral" button. Then find the area of the
rectangle.
The answer of the given question based on the finding the area of rectangle is QRST is approximately 116.56 square units.
What is Area?Area is a measure of the size or extent of a two-dimensional surface, like a shape or a region on a flat surface. It is typically measured in square units, like square meters (m²), square feet (ft²), or square centimeters (cm²).
To calculate the area of a shape, you need to know its dimensions and shape. The formula for finding the area of a shape depends on its shape. Area is an important concept in mathematics and geometry, and it is used in many real-life applications, such as measuring the size of land, determining the amount of material needed for a project, and calculating the volume of a container or room.
To find the area of the rectangle, we can use the distance formula to find the lengths of its sides. Then, we can use the formula for the area of a rectangle, which is the product of its length and width.
Length QR = √[(6 - (-4))² + (3 - (-3))²] = √[10² + 6²] = √(136)
Length ST = √[(9 - (-1))² + (-2 - (-8))²] = √[10² + 6²] = √(136)
Width QS = √[(6 - (-1))² + (3 - (-8))²] = √[7² + 11²] = √(170)
Width RT = √[(-4 - 9)² + (-3 - (-2))²] = √[13² + 1] = √(170)
Therefore, the area of the rectangle is:
Area = Length x Width = √(136) x √(170) = 116.56 (rounded to two decimal places)
So, the area of the rectangle QRST is approximately 116.56 square units.
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Write the equation of the circle graphed below
Answer: It basically just Pi
Step-by-step explanation:
Pi equals 27/3 right, so you say pi = 27/3
That's basically it!
Find the radian measure of all angles whose cosine is -0.89
(One of the answers is 3.62, but shouldn’t we add pi instead of minus 2pi) (I have a test tomorrow please answer if you know!)
Check the picture below.
so going on the 1st revolution from 0 to 2π, we get those two angles you see in the picture, the cosine is positive on I and IV Quadrants, whilst negative on the II and III Quadrants, so any angles with that negative cosine will be only on those two Quadrants.
when getting an angle or a trig function, make sure your calculator is in the correct mode, because plugging in cos(-0.89) in Degree mode, it's going to assume you mean to find the cosine of -0.89 Degrees, and same will happen the other way around, so make sure your mode is correct, so the input is understood correctly.
the cos⁻¹ function has a range of 0 to π, so any angles it's going to spit out will be on those Quadrants, we get the one in the III Quadrant in the picture by simply using the reference angle from the II Quadrant, the reference angle is about 0.47359 radians, so if we add that to π, we get that many radians.
now, if we want to include All revolutions around the circle, we can say
[tex]\approx ~~ \begin{cases} 2.668+2\pi n\\\\ 3.615+2\pi n \end{cases}\qquad n\in \mathbb{Z};n\geqslant 0[/tex]
The size of a certain insect population is given by P(t) = 100e^.03t. where t is measured in days. a. How many insects were present initially? b. Give a differential equation satisfied by P(t). c. At what time will the population double?
d. At what time will the population equal 300?
We have that, The size of a given insect population is given by [tex]P(t) = 100e^{.03t}[/tex]. Where t is measured in days, we obtain the following:
a. Initially there were 100 insectsb. The differential equation satisfied by P(t) is: [tex]dP(t)/dt = 3e^{(.03t)}[/tex]c. The population will approximately double in t = 23.10 days.d. The population will equal 300 approximately in t = 36.32 days.a. To find the initial size of the insect population, we need to evaluate P(t) at t=0:
[tex]P(0) = 100e^(.03 * 0)P(0) = 100e^0\\P(0) = 100(1)\\P(0) = 100\\[/tex]
Therefore, initially there were 100 insects.
b. To find the differential equation satisfied by P(t), we need to find the derivative of P(t) with respect to time (t):
[tex]dP(t)/dt = d(100e^{(.03t))}/dt\\dP(t)/dt = 100(.03)e^{(.03t)}[/tex]
Then, the differential equation satisfied by P(t) is:
[tex]dP(t)/dt = 3e^{(.03t)}[/tex]
c. To find the time when the population doubles, we need to find t when P(t) = 200:
[tex]200 = 100e^{(.03t)}\\2 = e^{(.03t)}\\[/tex]
Now, take the natural logarithm (ln) of both sides:
[tex]ln(2) = ln(e^{(.03t)})\\ln(2) = .03t[/tex]
Solve for t:
[tex]t = ln(2)/.03 \approx 23.10.[/tex]
Therefore, the population will approximately double in t = 23.10 days.
d. To find the time when the population equals 300, we must solve for t at P(t) = 300:
[tex]300 = 100e^{(.03t)}\\3 = e^{(.03t)}[/tex]
Take the natural logarithm (ln) of both sides:
[tex]ln(3) = ln(e^{(.03t)})\\ln(3) = .03t[/tex]
Solve for t:
[tex]t = ln(3)/.03 \approx 36.32.[/tex]
Therefore, the population will equal approximately 300 at t = 36.32 days.
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two angles directly across from each other on intersecting lines Is called?
PLESE HURRY 50 POINTSSS
Find the volume of the composite solid. Round your answer to the nearest tenth.
Answer:
about 1 cubic ft i think
Step-by-step explanation:
In the 2021 football season, the Riverside Red Dragon's manager gathered the points scored by the team during the regular season games. The following data set shows the values collected by the manager.
{17, 0, 28, 17, 17, 20, 28, 11, 17, 22, 24, 33, 20, 31, 20, 3}
Which of the following stem-and-leaf plots correctly graphs the data set?
2021 Riverside Red Dragon Football Season Points
0 0 3
1 1 7 7 7 7
2 0 0 0 2 4 8 8
3 1 3
Key 1|1 = 11
2021 Riverside Red Dragon Football Season Points
0 0 3
1 1 7
2 0 2 4 8
3 1 3
Key 1|1 = 11
2021 Riverside Red Dragon Football Season Points
1 1 7
2 0 2 4 8
3 0 1 3
Key 1|1 = 11
2021 Riverside Red Dragon Football Season Points
1 1 7 7 7 7
2 0 0 0 2 4 8 8
3 0 1 3
Key 1|1 = 11
2021 Riverside Red Dragon Football Season Points
0 0 3
1 1 7 7 7 7
2 0 0 0 2 4 8 8
3 1 3
Key 1|1 = 11
This stem-and-leaf plot correctly graphs the data set.
Irene needs to make at least 25 dinners for a party,
including chicken dinners and vegetarian dinners.
She has $250 to spend. Chicken dinners cost $8.75
each and vegetarian dinners cost $5.50 each.
• c represents the number of chicken dinners.
v represents the number of vegetarian dinners.
Write a system of inequalities that represents Irene's
constraints.
First inequality,
Second inequality
The system of inequalities that represents Irene's constraints is:
c + v ≥ 25 (She needs to make at least 25 dinners) 8.75c + 5.50v ≤ 250.
The problem presents two types of dinners: chicken and vegetarian. Irene needs to make at least 25 dinners, so the total number of dinners must be greater than or equal to 25. The cost of chicken dinners and vegetarian dinners is also given, along with the total budget of $250. These constraints can be translated into mathematical inequalities using variables c and v to represent the number of chicken and vegetarian dinners, respectively.
The first inequality represents the constraint that the total number of dinners must be greater than or equal to 25:
c + v ≥ 25
The second inequality represents the constraint that the cost of the dinners cannot exceed the total budget of $250:
8.75c + 5.50v ≤ 250
Together, these two inequalities form a system of inequalities that represents Irene's constraints. The solution to this system will give the possible values of c and v that satisfy both constraints.
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What is the verbal description of y=2/5x. WHO EVER ANSWERS WILL GET BRAINLIEST!!
The equation describes a linear relationship between two variables, where the output of the equation, y, is equal to two fifths of the input, x.
To calculate the value of y in the equation y=2/5x, we need to start by identifying the value of x. Once that is known, the value of y can be calculated by substituting x into the equation. For example, if x=10, then we can substitute 10 into the equation to get y=2/5 x 10. To solve this, we can multiply 2/5 by 10 to get 2x10/5, which simplifies to 20/5. Since dividing by 5 is the same as multiplying by 1/5, we can simplify this to 20/5 = 4. Therefore, when x=10, y=4.
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the complete question is
What is the verbal description of y=2/5x and identify the value of x
Subtract
−
5
x
2
−
9
x
−
5
−5x
2
−9x−5 from
4
x
2
−
5
x
+
3
4x
2
−5x+3.
Answer:
To subtract −5x² − 9x − 5 from 4x² − 5x + 3, we need to subtract the coefficients of each term with the same degree.
(4x² - 5x + 3) - (-5x² - 9x - 5)
4x² - 5x + 3 + 5x² + 9x + 5
9x² + 4x + 8
To subtract -5x² - 9x - 5 from 4x² - 5x + 3, you subtract each corresponding term, which results in 9x² + 4x + 8.
Explanation:If you would like to subtract -5x² - 9x - 5 from 4x² - 5x + 3, you need to subtract every term separately. So, the steps will look like the following:
subtract -5x² from 4x² which equals 9x²subtract -9x from -5x which equals 4xsubtract -5 from 3 which equals 8So, 4x² - 5x + 3 - (-5x² - 9x - 5) = 9x² + 4x + 8.
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PLS HELP Write the equation that is parallel to y = -3x + 7 and passes through the point (2, -1). Write your answer in slope intercept form.
3.12 freeway level of service. a four-lane urban freeway (two lanes in each direction) was built in part using an abandoned canal. the design speed is 60 mph. the lanes are 10 feet wide without any shoulder on either side. if the traffic flow is 2500 vph with 25 percent trucks in the peak hour, what level of service is the freeway operation if there is an interchange every 0.33 mile? what is the density? g
The level of service for the urban freeway is level E, and the density is 61.25 vehicles per mile.
To determine the level of service for the urban freeway, we can use the Highway Capacity Manual (HCM) method, which considers factors such as traffic volume, speed, and density.
Based on the given traffic flow of 2500 vehicles per hour and a design speed of 60 mph, the capacity of the four-lane freeway can be calculated as 8000 vehicles per hour. Using this value, along with the interchange spacing of 0.33 miles, we can determine the density of the freeway, which is 61.25 vehicles per mile.
Using the HCM method, the level of service is determined by comparing the freeway density to the capacity, which results in a level of service of E. Level E indicates that the freeway is operating at or near its capacity, and delays are frequent.
In conclusion, the urban freeway has a level of service classified as E, and density of the freeway, which is the number of vehicles per mile, is calculated to be 61.25.
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After identifying each of the mistakes that Ben and luna made, find the correct area of the circle. Round your answer to the nearest hundredth.
The area of the circle is 63.585 sq. cm.
What is area of circle?An object's area is how much space it takes up in two dimensions. It is the measurement of the quantity of unit squares that completely cover the surface of a closed figure. The square unit, which is frequently expressed as square inches, square feet, etc., is the accepted unit of area. Each shape's area may be calculated by counting how many unit squares will fit inside of it.
Given that the diameter of the circle is 9 cm
The radius = 9/2 = 4.5 cm
The area of the circle is:
A = πr²
Substitute the values:
A = (3.14)(4.5)²
A = 63.585 sq. cm.
Hence, the area of the circle is 63.585 sq. cm.
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The complete question is:
Follow the steps and match the order (this is not multiple choice question. Only use 3 steps)
12x-6=10
6x-3=5
6x=8
X=4/3
A. Add 3 to both sides
B. Divide both sides by 2
C. Subtract 3 from both sides
D. Multiply both sides by 1/6
E. Add 6 to both sides
Suppose that an individual has a body fat percentage of 17.3% and weighs 163 pounds. How many pounds of his weight is made up of fat? Round your answer
to the nearest tenth
After answering the presented question, we can conclude that Hence, fat expression accounts for around 28.2 pounds of the individual's weight.
what is expression ?An expression in mathematics is a collection of representations, digits, and conglomerates that mimic a statistical correlation or regularity. A real number, a mutable, or a combination of the two can be used as an expression. Mathematical operators include addition, subtraction, rapid spread, division, and exponentiation. Expressions are often used in arithmetic, mathematics, and form. They are employed in the representation of mathematical formulas, the solving of equations, and the simplification of mathematical relationships.
To establish how many pounds of an individual's weight are made up of fat, we must first determine the individual's body fat weight.
We may accomplish this by multiplying the individual's weight by the percentage of body fat (in decimal form):
Body fat weight = 163 pounds x 0.173 = 28.199 pounds
Hence, fat accounts for around 28.2 pounds of the individual's weight.
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The value of 3/3 + 4/
2 =. (5 points)
HELP PLEASE ITS DO IN 15 MIN!!!!
Answer: 3/3 +4/2=3
Step-by-step explanation:
given: 3/3 + 4/2
3/3+ 4/2 =1 +2
=3
Answer:
3/3+4/2=3
Step-by-step explanation:
3/3+4/2
=1+2
=3
A number added to twice another number is −8. The sum of the two numbers is −2. What is the lesser of these two numbers?
Answer:
The two numbers are 4 and -6. Since -6 is less than 4, the lesser of these two numbers is -6.
The lesser number is -6.
What is expression?Mathematical expressions consist of at least two numbers or variables, at least one arithmetic operation, and a statement. It's possible to multiply, divide, add, or subtract with this mathematical operation.
Let's call the two numbers x and y. We know that:
x + y = -2 (the sum of the two numbers is -2)x + 2y = -8 (a number added to twice another number is -8)We can solve this system of equations using substitution. Solving the first equation for x, we get:
x = -2 - y
We can substitute this expression for x in the second equation:
-2 - y + 2y = -8
Simplifying this equation, we get:
y = -6
Now we can use the first equation to find x:
x + y = -2
x + (-6) = -2
x = 4
So the two numbers are -6 and 4, and the lesser of the two numbers is -6.
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A northbound car is going 12 miles per hour faster than a southbound car.
The cars are 276 miles apart 3 hours after passing each other on the interstate.
What is the speed of the northbound car?
Check the picture below.
so the cars are 276 miles apart, let's say the southbound car is going "r" mph fast, that means the northbound car is going "r + 12" mph fast as you see in the picture, so at some time they meet, hmmm wait, we know they're 3 hours apart from passing each other, so that means by the time they meet, 3 hours have passed, so by the time that happens, the northbound car has been traveling for 3 hours and the southbound car has been traveling for 3 hours as well.
Since we know they're 276 miles apart, let's say by the time they meet the southbound car has covered "d" miles, so the northbound car has covered the slack then, that is "276 - d" miles.
[tex]{\Large \begin{array}{llll} \underset{distance}{d}=\underset{rate}{r} \stackrel{time}{t} \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Southbound&d&r&3\\ Northbound&276-d&r+12&3 \end{array}\hspace{5em} \begin{cases} d=(r)(3)\\\\ 276-d=(r+12)(3) \end{cases} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{\textit{using the 1st equation}}{d=3r}\hspace{5em}\stackrel{\textit{substituting on the 2nd equation}}{276-(3r)~~ = ~~3(r+12)} \\\\\\ 276-3r=3r+36\implies 240-3r=3r\implies 240=6r \\\\\\ \cfrac{240}{6}=r\implies \stackrel{ Southbound~car }{40=r}\hspace{5em}\underset{ Northbound~car }{\stackrel{ 40~~ + ~~12 }{\text{\LARGE 52}}}[/tex]