You place ice cream on a metal surface that is at a room temperature of 25 degrees Celsius. Which will happen when the ice cream and metal surface interact?(1 point)
Responses
Heat is transferred from the metal surface to the ice cream, and the ice cream melts.
Heat is transferred from the metal surface to the ice cream, and the ice cream melts.
Heat is transferred from the ice cream to the metal surface, and the ice cream melts.
Heat is transferred from the ice cream to the metal surface, and the ice cream melts.
Heat is transferred from the ice cream to the air, and the ice cream melts.
Heat is transferred from the ice cream to the air, and the ice cream melts.
Heat is transferred from the ice cream and metal surface to the air, and the ice cream melts.
23 answers
Heat is transferred from the metal surface to the ice cream, and the ice cream melts.
An iron block with a temperature of 250 degrees Celcius is placed into a container of water with a temperature of 50 degrees Celcius. Which statement best describes what will be observed in this experiment?
Heat will flow from the water to the iron block until both are at the temperature of the container.
Heat will flow from the water to the iron block until both are at the temperature of the container.
Heat will flow from the iron block to the water until both are 50 degrees Celcius.
Heat will flow from the iron block to the water until both are 50 degrees Celcius.
Heat will flow from the iron block to the water until both are 250 degrees Celcius.
Heat will flow from the iron block to the water until both are 250 degrees Celcius.
Heat will flow from the iron block to the block to the water until both reach a temperature somewhere between 500 degrees Celcius and 250 degrees Celcius.
Heat will flow from the water to the iron block until both are at the temperature of the container.
Heat will flow from the water to the iron block until both are at the temperature of the container.
Heat will flow from the iron block to the water until both are 50 degrees Celcius.
Heat will flow from the iron block to the water until both are 50 degrees Celcius.
Heat will flow from the iron block to the water until both are 250 degrees Celcius.
Heat will flow from the iron block to the water until both are 250 degrees Celcius.
Heat will flow from the iron block to the block to the water until both reach a temperature somewhere between 500 degrees Celcius and 250 degrees Celcius.
Heat will flow from the iron block to the water until both are at the temperature of the container.
A student drops substance A into substance B. What will happen?(1 point)
Responses
Both substances will reach 50 degrees Celsius due to heat exchange.
Both substances will reach 50 degrees Celsius due to heat exchange.
Both substances will reach 25 degrees Celsius due to heat exchange.
Both substances will reach 25 degrees Celsius due to heat exchange.
Substance A will become warmer as heat flows from substance A to substance B.
Substance A will become warmer as heat flows from substance A to substance B.
After some time passes, the substances will reach the same temperature.
Responses
Both substances will reach 50 degrees Celsius due to heat exchange.
Both substances will reach 50 degrees Celsius due to heat exchange.
Both substances will reach 25 degrees Celsius due to heat exchange.
Both substances will reach 25 degrees Celsius due to heat exchange.
Substance A will become warmer as heat flows from substance A to substance B.
Substance A will become warmer as heat flows from substance A to substance B.
After some time passes, the substances will reach the same temperature.
After some time passes, the substances will reach the same temperature.
When you add cold milk and hot chocolate, they will slowly blend as the cooler liquid mixes with the warmer liquid. Eventually, the heat will be spread out evenly through the entire liquid. Which statement best explains why this happens?(1 point)
Responses
The molecules of the two liquids form chemical bonds to spread energy.
The molecules of the two liquids form chemical bonds to spread energy.
Heat always moves from areas of higher temperature to areas of lower temperature.
Heat always moves from areas of higher temperature to areas of lower temperature.
The liquids attract each other because the densities of the liquids are the same.
The liquids attract each other because the densities of the liquids are the same.
Cold liquids rise and hot liquids sink making the liquids mix.
Responses
The molecules of the two liquids form chemical bonds to spread energy.
The molecules of the two liquids form chemical bonds to spread energy.
Heat always moves from areas of higher temperature to areas of lower temperature.
Heat always moves from areas of higher temperature to areas of lower temperature.
The liquids attract each other because the densities of the liquids are the same.
The liquids attract each other because the densities of the liquids are the same.
Cold liquids rise and hot liquids sink making the liquids mix.
Heat always moves from areas of higher temperature to areas of lower temperature.
Which equation has the same first step to solve as this equation: 2x+12−3x=27(1 point)
Responses
4x−17+2x=9
2x+12=27
2x+12=−3x−27
2(x−5)+2x=10
Responses
4x−17+2x=9
2x+12=27
2x+12=−3x−27
2(x−5)+2x=10
2x+12=−3x−27
Using the Distributive Property as a good first step to solving the equation 7(5x+2)=−4(6−5x), you could simplify this equation to get which of these choices?(1 point)
35x+14=−24+20x
5x+14=6+20x
35x+14=−24−20x
35x+2=−24−5x
35x+2=24−20x
35x+14=−24+20x
5x+14=6+20x
35x+14=−24−20x
35x+2=−24−5x
35x+2=24−20x
35x+14=−24+20x
After combining like terms to simplify the equation 3−15x+24+16x=4x−24−4x, what would be the next best step to finish solving?(1 point)
Responses
Subtract x from both sides of the equation.
Subtract x from both sides of the equation.
Divide both sides of the equation by 15.
Divide both sides of the equation by 15.
Subtract 24 from both sides of the equation.
Subtract 24 from both sides of the equation.
Subtract 27 from both sides of the equation.
Subtract 27 from both sides of the equation.
Add x to both sides of the equation.
Add x to both sides of the equation.
Responses
Subtract x from both sides of the equation.
Subtract x from both sides of the equation.
Divide both sides of the equation by 15.
Divide both sides of the equation by 15.
Subtract 24 from both sides of the equation.
Subtract 24 from both sides of the equation.
Subtract 27 from both sides of the equation.
Subtract 27 from both sides of the equation.
Add x to both sides of the equation.
Add x to both sides of the equation.
Subtract 24 from both sides of the equation.
The speed limit sign on the expressway says that the speed limit is 55 mph. Write an inequality that represents all speeds you can legally drive on the expressway. (1 point)
The inequality representing all speeds you can legally drive on the expressway would be: v ≤ 55, where v represents the speed in mph.
You originally have x colored pencils in your pencil box. You share them equally between you and your friend. You add in 3 more colored pencils to your own box after sharing them. This gives you a total of 15 pencils.
a. Write an equation and show all of your steps for solving x (the original number of colored pencils in your box).
b. If you wanted to have at least 20 colored pencils left in your box, how many colored pencils would you have needed originally?
c. Write an inequality and show all steps for solving for x.
a. Write an equation and show all of your steps for solving x (the original number of colored pencils in your box).
b. If you wanted to have at least 20 colored pencils left in your box, how many colored pencils would you have needed originally?
c. Write an inequality and show all steps for solving for x.
a. Let's solve for x, the original number of colored pencils in your box.
Step 1: You share the pencils equally with your friend, so each of you gets (x/2) pencils.
Step 2: You add 3 more colored pencils to your own box, so you have (x/2) + 3 pencils.
Step 3: The total number of colored pencils you have is (x/2) + 3.
Step 4: According to the problem, the total number of colored pencils you have is 15.
So, we can set up the equation: (x/2) + 3 = 15.
Now, let's solve this equation for x.
(x/2) + 3 = 15
Subtract 3 from both sides:
(x/2) + 3 - 3 = 15 - 3
(x/2) = 12
Multiply both sides by 2 to isolate x:
2 * (x/2) = 12 * 2
x = 24
So, the original number of colored pencils in your box (x) is 24.
b. If you wanted to have at least 20 colored pencils left in your box, let's solve for the original number of colored pencils needed.
Step 1: Let's assume the original number of colored pencils as y.
Step 2: You share the pencils equally with your friend, so each of you gets (y/2) pencils.
Step 3: You add 3 more colored pencils to your own box, so you have (y/2) + 3 pencils.
Step 4: You want to have at least 20 colored pencils left, so we can set up the inequality: (y/2) + 3 ≥ 20.
Now, let's solve this inequality for y.
(y/2) + 3 ≥ 20
Subtract 3 from both sides:
(y/2) + 3 - 3 ≥ 20 - 3
(y/2) ≥ 17
Multiply both sides by 2 to isolate y:
2 * (y/2) ≥ 17 * 2
y ≥ 34
So, the original number of colored pencils needed (y) would be 34 or any number greater than or equal to 34.
c. The inequality representing the original number of colored pencils in your box would be: x ≥ 34, where x represents the original number of colored pencils.
Step 1: You share the pencils equally with your friend, so each of you gets (x/2) pencils.
Step 2: You add 3 more colored pencils to your own box, so you have (x/2) + 3 pencils.
Step 3: The total number of colored pencils you have is (x/2) + 3.
Step 4: According to the problem, the total number of colored pencils you have is 15.
So, we can set up the equation: (x/2) + 3 = 15.
Now, let's solve this equation for x.
(x/2) + 3 = 15
Subtract 3 from both sides:
(x/2) + 3 - 3 = 15 - 3
(x/2) = 12
Multiply both sides by 2 to isolate x:
2 * (x/2) = 12 * 2
x = 24
So, the original number of colored pencils in your box (x) is 24.
b. If you wanted to have at least 20 colored pencils left in your box, let's solve for the original number of colored pencils needed.
Step 1: Let's assume the original number of colored pencils as y.
Step 2: You share the pencils equally with your friend, so each of you gets (y/2) pencils.
Step 3: You add 3 more colored pencils to your own box, so you have (y/2) + 3 pencils.
Step 4: You want to have at least 20 colored pencils left, so we can set up the inequality: (y/2) + 3 ≥ 20.
Now, let's solve this inequality for y.
(y/2) + 3 ≥ 20
Subtract 3 from both sides:
(y/2) + 3 - 3 ≥ 20 - 3
(y/2) ≥ 17
Multiply both sides by 2 to isolate y:
2 * (y/2) ≥ 17 * 2
y ≥ 34
So, the original number of colored pencils needed (y) would be 34 or any number greater than or equal to 34.
c. The inequality representing the original number of colored pencils in your box would be: x ≥ 34, where x represents the original number of colored pencils.
You originally have x colored pencils in your pencil box. You share them equally between you and your friend. You add in 3 more colored pencils to your own box after sharing them. This gives you a total of 15 pencils.
a. Write an equation and show all of your steps for solving x (the original number of colored pencils in your box).
a. Write an equation and show all of your steps for solving x (the original number of colored pencils in your box).
Sure, let's solve for x, the original number of colored pencils in your box.
Step 1: You share the pencils equally with your friend, so each of you gets (x/2) pencils.
Step 2: You add 3 more colored pencils to your own box, so you have (x/2) + 3 pencils.
Step 3: The total number of colored pencils you have is (x/2) + 3.
Step 4: According to the problem, the total number of colored pencils you have is 15.
So, we can set up the equation: (x/2) + 3 = 15.
To solve the equation:
Step 5: Subtract 3 from both sides of the equation.
(x/2) + 3 - 3 = 15 - 3,
(x/2) = 12.
Step 6: Multiply both sides of the equation by 2 to isolate x.
2 * (x/2) = 12 * 2,
x = 24.
Therefore, the original number of colored pencils in your box (x) is 24.
Step 1: You share the pencils equally with your friend, so each of you gets (x/2) pencils.
Step 2: You add 3 more colored pencils to your own box, so you have (x/2) + 3 pencils.
Step 3: The total number of colored pencils you have is (x/2) + 3.
Step 4: According to the problem, the total number of colored pencils you have is 15.
So, we can set up the equation: (x/2) + 3 = 15.
To solve the equation:
Step 5: Subtract 3 from both sides of the equation.
(x/2) + 3 - 3 = 15 - 3,
(x/2) = 12.
Step 6: Multiply both sides of the equation by 2 to isolate x.
2 * (x/2) = 12 * 2,
x = 24.
Therefore, the original number of colored pencils in your box (x) is 24.
If you wanted to have at least 20 colored pencils left in your box, how many colored pencils would you have needed originally?
If you want to have at least 20 colored pencils left in your box after giving half of them to your friend and adding 3 more, we can set up the following inequality:
(x/2) + 3 ≥ 20
To solve for x, the original number of colored pencils in your box:
Step 1: Subtract 3 from both sides of the inequality:
(x/2) + 3 - 3 ≥ 20 - 3
(x/2) ≥ 17
Step 2: Multiply both sides of the inequality by 2 (since you're dividing by 2 on the left side) to isolate x:
2 * (x/2) ≥ 17 * 2
x ≥ 34
Therefore, to have at least 20 colored pencils left in your box, you would have needed originally 34 colored pencils or any number greater than or equal to 34.
(x/2) + 3 ≥ 20
To solve for x, the original number of colored pencils in your box:
Step 1: Subtract 3 from both sides of the inequality:
(x/2) + 3 - 3 ≥ 20 - 3
(x/2) ≥ 17
Step 2: Multiply both sides of the inequality by 2 (since you're dividing by 2 on the left side) to isolate x:
2 * (x/2) ≥ 17 * 2
x ≥ 34
Therefore, to have at least 20 colored pencils left in your box, you would have needed originally 34 colored pencils or any number greater than or equal to 34.
The base of a triangle is four less than three times the height. The area of the triangle is 38 square centimeters. A=(b×h)2(3 points)
An expression for the base of the triangle:
Equation for the area of the triangle, substituting in for the base:
The first step in solving for h:
An expression for the base of the triangle:
Equation for the area of the triangle, substituting in for the base:
The first step in solving for h:
To find the expression for the base of the triangle, we can let h represent the height of the triangle.
The base of the triangle is four less than three times the height. So the expression for the base can be written as:
base = 3h - 4
Next, let's write the equation for the area of the triangle, substituting in for the base:
Area = (base * height) / 2
Substituting the expression for base, the equation for the area becomes:
Area = ((3h - 4) * h) / 2
Now, let's determine the first step in solving for h.
To find the first step in solving for h, we need to set up the equation using the given information about the area of the triangle.
We know that the area of the triangle is 38 square centimeters, so we can set up the equation:
38 = ((3h - 4) * h) / 2
This equation can then be used to find the first step in solving for h.
The base of the triangle is four less than three times the height. So the expression for the base can be written as:
base = 3h - 4
Next, let's write the equation for the area of the triangle, substituting in for the base:
Area = (base * height) / 2
Substituting the expression for base, the equation for the area becomes:
Area = ((3h - 4) * h) / 2
Now, let's determine the first step in solving for h.
To find the first step in solving for h, we need to set up the equation using the given information about the area of the triangle.
We know that the area of the triangle is 38 square centimeters, so we can set up the equation:
38 = ((3h - 4) * h) / 2
This equation can then be used to find the first step in solving for h.
1 answer