To solve for x, we need to isolate it on one side of the equation.
Given:
x + 35 = -13
To isolate x, we need to subtract 35 from both sides of the equation:
x + 35 - 35 = -13 - 35
Simplifying the equation:
x = -13 - 35
Combining the terms on the right side of the equation:
x = -48
Therefore, the solution to the equation x + 35 = -13 is x = -48.
x+35=-13
37 answers
x+1/10=29/90
To solve the equation x + 1/10 = 29/90, we need to isolate the variable x.
Step 1: Start by subtracting 1/10 from both sides of the equation:
x + 1/10 - 1/10 = 29/90 - 1/10
x = 29/90 - 1/10
Step 2: Find a common denominator for the fractions on the right side:
The common denominator of 90 and 10 is 90.
Step 3: Rewrite the fractions with the common denominator:
x = (29 * 10)/(90 * 10) - (1 * 90)/(10 * 90)
x = 290/900 - 90/900
Step 4: Combine the fractions on the right side of the equation:
x = (290 - 90)/900
x = 200/900
Step 5: Simplify the fraction:
x = 2/9
Therefore, the solution to the equation x + 1/10 = 29/90 is x = 2/9.
Step 1: Start by subtracting 1/10 from both sides of the equation:
x + 1/10 - 1/10 = 29/90 - 1/10
x = 29/90 - 1/10
Step 2: Find a common denominator for the fractions on the right side:
The common denominator of 90 and 10 is 90.
Step 3: Rewrite the fractions with the common denominator:
x = (29 * 10)/(90 * 10) - (1 * 90)/(10 * 90)
x = 290/900 - 90/900
Step 4: Combine the fractions on the right side of the equation:
x = (290 - 90)/900
x = 200/900
Step 5: Simplify the fraction:
x = 2/9
Therefore, the solution to the equation x + 1/10 = 29/90 is x = 2/9.
y/5=15
To solve the equation y/5 = 15, we need to isolate the variable y.
Step 1: Multiply both sides of the equation by 5 to get rid of the fraction.
(y/5) * 5 = 15 * 5
y = 75
Therefore, the solution to the equation y/5 = 15 is y = 75.
Step 1: Multiply both sides of the equation by 5 to get rid of the fraction.
(y/5) * 5 = 15 * 5
y = 75
Therefore, the solution to the equation y/5 = 15 is y = 75.
0.9v=15.03
To solve the equation 0.9v = 15.03, we need to isolate the variable v.
Step 1: Divide both sides of the equation by 0.9 to get v alone.
(0.9v) / 0.9 = 15.03 / 0.9
v = 16.7
Therefore, the solution to the equation 0.9v = 15.03 is v = 16.7.
Step 1: Divide both sides of the equation by 0.9 to get v alone.
(0.9v) / 0.9 = 15.03 / 0.9
v = 16.7
Therefore, the solution to the equation 0.9v = 15.03 is v = 16.7.
a box of candy at the movie theater cost $4.50 and a drink costs $x the cost of the candy and drinks together is $6.75 complete and solve the equation that represents the cost of the drink
x+4.50= ------
x=------
x+4.50= ------
x=------
To represent the cost of the drink in the equation, we already know that the cost of the box of candy is $4.50. As given, the drink costs $x. Therefore, the total cost of the candy and drink together is $6.75.
The equation that represents the cost of the drink is:
x + 4.50 = 6.75
To solve for x, we need to isolate it on one side of the equation:
Subtract 4.50 from both sides:
x + 4.50 -4.50 = 6.75 - 4.50
Simplifying the equation:
x = 2.25
Therefore, the cost of the drink represented by x is $2.25.
The equation that represents the cost of the drink is:
x + 4.50 = 6.75
To solve for x, we need to isolate it on one side of the equation:
Subtract 4.50 from both sides:
x + 4.50 -4.50 = 6.75 - 4.50
Simplifying the equation:
x = 2.25
Therefore, the cost of the drink represented by x is $2.25.
amelia backed some cookies she was able to give 5 cookies each to 8 friends without having any cookies left over how could you set up this problem as an algebraic equation to find out how many cookies Amelia backed in total
Let's represent the total number of cookies that Amelia baked as "x".
Since Amelia gave 5 cookies to each of her 8 friends, the total number of cookies given away would be 8 multiplied by 5, which is 40 cookies.
According to the problem, there were no cookies left over after giving away the cookies. So, we can set up the equation:
x - 40 = 0
This equation tells us that the total number of cookies baked, represented by "x", minus the 40 cookies given away should be equal to zero cookies left over.
To solve for x, we can add 40 to both sides:
x - 40 + 40 = 0 + 40
This simplifies to:
x = 40
Therefore, Amelia baked a total of 40 cookies.
Since Amelia gave 5 cookies to each of her 8 friends, the total number of cookies given away would be 8 multiplied by 5, which is 40 cookies.
According to the problem, there were no cookies left over after giving away the cookies. So, we can set up the equation:
x - 40 = 0
This equation tells us that the total number of cookies baked, represented by "x", minus the 40 cookies given away should be equal to zero cookies left over.
To solve for x, we can add 40 to both sides:
x - 40 + 40 = 0 + 40
This simplifies to:
x = 40
Therefore, Amelia baked a total of 40 cookies.
can you show every thin in numbers
Certainly! Let's represent the total number of cookies that Amelia baked as "x".
Amelia gave 8 friends 5 cookies each, so the total number of cookies given away is 8 * 5 = 40 cookies.
We can set up the equation:
x - 40 = 0
To solve for x, we can add 40 to both sides:
x - 40 + 40 = 0 + 40
This simplifies to:
x = 40
Therefore, Amelia baked a total of 40 cookies.
Amelia gave 8 friends 5 cookies each, so the total number of cookies given away is 8 * 5 = 40 cookies.
We can set up the equation:
x - 40 = 0
To solve for x, we can add 40 to both sides:
x - 40 + 40 = 0 + 40
This simplifies to:
x = 40
Therefore, Amelia baked a total of 40 cookies.
john is cooking cupcakes the recipe calls for 3 3/10 cups of flour . he accidentally put in 3 5/8 cups how many extra cups did he put in
To find out how many extra cups of flour John put in, we need to subtract the required amount of flour from the amount he actually put in.
The required amount of flour for the recipe is 3 3/10 cups.
The amount John actually put in is 3 5/8 cups.
To subtract these two amounts, we need to make sure the fractions have a common denominator.
Converting 3 3/10 to an improper fraction:
3 3/10 = (3 * 10 + 3) / 10 = 33/10
Converting 3 5/8 to an improper fraction:
3 5/8 = (3 * 8 + 5) / 8 = 29/8
Now, we can subtract the two amounts:
29/8 - 33/10
To subtract fractions with different denominators, we need to find a common denominator, which is 8 * 10 = 80.
Rewriting the fractions with the common denominator:
(29/8) * (10/10) - (33/10) * (8/8)
290/80 - 264/80
Now we can subtract the fractions:
290/80 - 264/80 = (290 - 264) / 80 = 26/80
Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor:
26/80 = (26/2) / (80/2) = 13/40
Therefore, John put in an extra 13/40 cups of flour.
The required amount of flour for the recipe is 3 3/10 cups.
The amount John actually put in is 3 5/8 cups.
To subtract these two amounts, we need to make sure the fractions have a common denominator.
Converting 3 3/10 to an improper fraction:
3 3/10 = (3 * 10 + 3) / 10 = 33/10
Converting 3 5/8 to an improper fraction:
3 5/8 = (3 * 8 + 5) / 8 = 29/8
Now, we can subtract the two amounts:
29/8 - 33/10
To subtract fractions with different denominators, we need to find a common denominator, which is 8 * 10 = 80.
Rewriting the fractions with the common denominator:
(29/8) * (10/10) - (33/10) * (8/8)
290/80 - 264/80
Now we can subtract the fractions:
290/80 - 264/80 = (290 - 264) / 80 = 26/80
Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor:
26/80 = (26/2) / (80/2) = 13/40
Therefore, John put in an extra 13/40 cups of flour.
use only numbers
Certainly! Let's calculate the extra cups of flour John put in using only numbers.
The required amount of flour for the recipe is 3 3/10 cups.
The amount John actually put in is 3 5/8 cups.
To convert the mixed numbers to improper fractions:
3 3/10 = (3 * 10 + 3) / 10 = 33/10
3 5/8 = (3 * 8 + 5) / 8 = 29/8
To subtract the two amounts:
29/8 - 33/10
To subtract fractions with different denominators, we need to find a common denominator, which is 8 * 10 = 80.
Multiplying the numerator and denominator of the first fraction by 10:
(29/8) * (10/10) = 290/80
Multiplying the numerator and denominator of the second fraction by 8:
(33/10) * (8/8) = 264/80
Now we can subtract the fractions:
290/80 - 264/80 = (290 - 264) / 80 = 26/80
Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor:
26/80 = (26/2) / (80/2) = 13/40
Therefore, John put in an extra 13/40 cups of flour.
The required amount of flour for the recipe is 3 3/10 cups.
The amount John actually put in is 3 5/8 cups.
To convert the mixed numbers to improper fractions:
3 3/10 = (3 * 10 + 3) / 10 = 33/10
3 5/8 = (3 * 8 + 5) / 8 = 29/8
To subtract the two amounts:
29/8 - 33/10
To subtract fractions with different denominators, we need to find a common denominator, which is 8 * 10 = 80.
Multiplying the numerator and denominator of the first fraction by 10:
(29/8) * (10/10) = 290/80
Multiplying the numerator and denominator of the second fraction by 8:
(33/10) * (8/8) = 264/80
Now we can subtract the fractions:
290/80 - 264/80 = (290 - 264) / 80 = 26/80
Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor:
26/80 = (26/2) / (80/2) = 13/40
Therefore, John put in an extra 13/40 cups of flour.
make it not so complicated
Apologies for the confusion. Let's simplify the calculation.
The required amount of flour for the recipe is 3 3/10 cups.
The amount John actually put in is 3 5/8 cups.
To find the extra cups of flour John put in:
3 5/8 - 3 3/10
To subtract mixed numbers, we need to first find a common denominator.
The common denominator of 8 and 10 is 40.
Converting the mixed numbers to fractions with the common denominator:
3 5/8 = 29/8
3 3/10 = 33/10
Subtracting the fractions:
29/8 - 33/10 = (29 * 5) / (8 * 5) - (33 * 4) / (10 * 4)
= 145/40 - 132/40
= (145 - 132) / 40
= 13/40
Therefore, John put in an extra 13/40 cups of flour.
The required amount of flour for the recipe is 3 3/10 cups.
The amount John actually put in is 3 5/8 cups.
To find the extra cups of flour John put in:
3 5/8 - 3 3/10
To subtract mixed numbers, we need to first find a common denominator.
The common denominator of 8 and 10 is 40.
Converting the mixed numbers to fractions with the common denominator:
3 5/8 = 29/8
3 3/10 = 33/10
Subtracting the fractions:
29/8 - 33/10 = (29 * 5) / (8 * 5) - (33 * 4) / (10 * 4)
= 145/40 - 132/40
= (145 - 132) / 40
= 13/40
Therefore, John put in an extra 13/40 cups of flour.
8y-5=27
To solve the equation 8y - 5 = 27, we need to isolate the variable y.
Step 1: Add 5 to both sides of the equation:
8y - 5 + 5 = 27 + 5
Simplifying the equation:
8y = 32
Step 2: Divide both sides of the equation by 8:
(8y) / 8 = 32 / 8
Simplifying the equation:
y = 4
Therefore, the solution to the equation 8y - 5 = 27 is y = 4.
Step 1: Add 5 to both sides of the equation:
8y - 5 + 5 = 27 + 5
Simplifying the equation:
8y = 32
Step 2: Divide both sides of the equation by 8:
(8y) / 8 = 32 / 8
Simplifying the equation:
y = 4
Therefore, the solution to the equation 8y - 5 = 27 is y = 4.
-2X+6=18
To solve the equation -2x + 6 = 18, we need to isolate the variable x.
Step 1: Subtract 6 from both sides of the equation:
-2x + 6 - 6 = 18 - 6
This simplifies to:
-2x = 12
Step 2: Divide both sides of the equation by -2 (the coefficient of x) to solve for x:
(-2x) / -2 = 12 / -2
Simplifying the equation:
x = -6
Therefore, the solution to the equation -2x + 6 = 18 is x = -6.
Step 1: Subtract 6 from both sides of the equation:
-2x + 6 - 6 = 18 - 6
This simplifies to:
-2x = 12
Step 2: Divide both sides of the equation by -2 (the coefficient of x) to solve for x:
(-2x) / -2 = 12 / -2
Simplifying the equation:
x = -6
Therefore, the solution to the equation -2x + 6 = 18 is x = -6.
A/3+5=3
To solve the equation A/3 + 5 = 3, we need to isolate the variable A.
Step 1: Subtract 5 from both sides of the equation:
A/3 + 5 - 5 = 3 - 5
This simplifies to:
A/3 = -2
Step 2: Multiply both sides of the equation by 3 (the denominator of A/3) to solve for A:
(A/3) * 3 = -2 * 3
Simplifying the equation:
A = -6
Therefore, the solution to the equation A/3 + 5 = 3 is A = -6.
Step 1: Subtract 5 from both sides of the equation:
A/3 + 5 - 5 = 3 - 5
This simplifies to:
A/3 = -2
Step 2: Multiply both sides of the equation by 3 (the denominator of A/3) to solve for A:
(A/3) * 3 = -2 * 3
Simplifying the equation:
A = -6
Therefore, the solution to the equation A/3 + 5 = 3 is A = -6.
x/9=3
To solve the equation x/9 = 3, we need to isolate the variable x.
Step 1: Multiply both sides of the equation by 9 to get rid of the fraction:
(x/9) * 9 = 3 * 9
This simplifies to:
x = 27
Therefore, the solution to the equation x/9 = 3 is x = 27.
Step 1: Multiply both sides of the equation by 9 to get rid of the fraction:
(x/9) * 9 = 3 * 9
This simplifies to:
x = 27
Therefore, the solution to the equation x/9 = 3 is x = 27.
at the grocery store ryan bought a loaf of bread and some apples the loaf of breads cost 3.95 and the apples cost 1.29 per pound if spent 10.40 for the loaf of bread and apples haw many pounds of apple did he buy set the problem and solve is
Let's represent the number of pounds of apples that Ryan bought as "x".
The cost of the loaf of bread is $3.95.
The cost of the apples is $1.29 per pound.
The total amount spent on the loaf of bread and apples is $10.40.
To set up the problem, we can create an equation:
3.95 + 1.29x = 10.40
This equation represents the cost of the loaf of bread ($3.95) plus the cost of the apples ($1.29 per pound multiplied by the number of pounds, x) equal to the total amount spent ($10.40).
To solve for x, we need to isolate it on one side of the equation.
Subtracting 3.95 from both sides:
3.95 - 3.95 + 1.29x = 10.40 - 3.95
Simplifying the equation:
1.29x = 6.45
To find the value of x, we divide both sides of the equation by 1.29:
(1.29x) / 1.29 = 6.45 / 1.29
Simplifying the equation:
x = 5
Therefore, Ryan bought 5 pounds of apples.
The cost of the loaf of bread is $3.95.
The cost of the apples is $1.29 per pound.
The total amount spent on the loaf of bread and apples is $10.40.
To set up the problem, we can create an equation:
3.95 + 1.29x = 10.40
This equation represents the cost of the loaf of bread ($3.95) plus the cost of the apples ($1.29 per pound multiplied by the number of pounds, x) equal to the total amount spent ($10.40).
To solve for x, we need to isolate it on one side of the equation.
Subtracting 3.95 from both sides:
3.95 - 3.95 + 1.29x = 10.40 - 3.95
Simplifying the equation:
1.29x = 6.45
To find the value of x, we divide both sides of the equation by 1.29:
(1.29x) / 1.29 = 6.45 / 1.29
Simplifying the equation:
x = 5
Therefore, Ryan bought 5 pounds of apples.
options
1.29x+395=10.40 ryan bought 5 pounds
3.95+10.40 ryan bought 4 pounds
10.40 - 1.29= 3.95x he bought 6.45 pounds
3.95-1.29x=10.40 he bought 5.16 pounds
1.29x+395=10.40 ryan bought 5 pounds
3.95+10.40 ryan bought 4 pounds
10.40 - 1.29= 3.95x he bought 6.45 pounds
3.95-1.29x=10.40 he bought 5.16 pounds
The correct option is:
1.29x + 3.95 = 10.40
To solve the equation, we will isolate the variable x.
Subtracting 3.95 from both sides:
1.29x + 3.95 - 3.95 = 10.40 - 3.95
Simplifying the equation:
1.29x = 6.45
To find the value of x, divide both sides of the equation by 1.29:
(1.29x) / 1.29 = 6.45 / 1.29
Simplifying the equation:
x = 5
Therefore, Ryan bought 5 pounds of apples.
1.29x + 3.95 = 10.40
To solve the equation, we will isolate the variable x.
Subtracting 3.95 from both sides:
1.29x + 3.95 - 3.95 = 10.40 - 3.95
Simplifying the equation:
1.29x = 6.45
To find the value of x, divide both sides of the equation by 1.29:
(1.29x) / 1.29 = 6.45 / 1.29
Simplifying the equation:
x = 5
Therefore, Ryan bought 5 pounds of apples.
jack won 65 lollipops playing basketball at the county fair at school he gave four to every student in his math class he only has 1 remaining how many students are in his
Let's represent the number of students in Jack's math class as "x".
Jack won 65 lollipops playing basketball at the county fair. He gave four lollipops to each student in his math class, and he has 1 lollipop remaining.
To set up the problem as an equation, we can multiply the number of students (x) by 4 (for the lollipops given to each student), and then add 1 for the lollipop remaining. This should equal the total number of lollipops won, which is 65.
4x + 1 = 65
To solve for x, we need to isolate it on one side of the equation.
Subtracting 1 from both sides:
4x + 1 - 1 = 65 - 1
Simplifying the equation:
4x = 64
To find the value of x, we divide both sides of the equation by 4:
(4x) / 4 = 64 / 4
Simplifying the equation:
x = 16
Therefore, there are 16 students in Jack's math class.
Jack won 65 lollipops playing basketball at the county fair. He gave four lollipops to each student in his math class, and he has 1 lollipop remaining.
To set up the problem as an equation, we can multiply the number of students (x) by 4 (for the lollipops given to each student), and then add 1 for the lollipop remaining. This should equal the total number of lollipops won, which is 65.
4x + 1 = 65
To solve for x, we need to isolate it on one side of the equation.
Subtracting 1 from both sides:
4x + 1 - 1 = 65 - 1
Simplifying the equation:
4x = 64
To find the value of x, we divide both sides of the equation by 4:
(4x) / 4 = 64 / 4
Simplifying the equation:
x = 16
Therefore, there are 16 students in Jack's math class.
only use numbers
Apologies for the previous response. Let's solve the problem using numbers.
Jack won 65 lollipops and gave four to each student in his math class. He has 1 lollipop remaining.
To find the number of students in his math class, we divide the remaining lollipop by the number of lollipops each student received:
1 / 4 = 0.25
Therefore, there are 0.25 or 1/4 of a student in Jack's math class.
Since it is not possible to have a fraction of a student, we can conclude that Jack's math class has 0 students.
Jack won 65 lollipops and gave four to each student in his math class. He has 1 lollipop remaining.
To find the number of students in his math class, we divide the remaining lollipop by the number of lollipops each student received:
1 / 4 = 0.25
Therefore, there are 0.25 or 1/4 of a student in Jack's math class.
Since it is not possible to have a fraction of a student, we can conclude that Jack's math class has 0 students.
jahan practices the drums for 45min on saturday and 25 min on most weekday last week he practiced 120 min how many weekdays did he practice write an equation in the form of px+q=r to represent the problem ---x----=
Let's represent the number of weekdays Jahan practiced the drums as "x".
On Saturday, Jahan practiced for 45 minutes.
On most weekdays, he practiced for 25 minutes each.
The total time Jahan practiced last week was 120 minutes.
To represent the problem as an equation in the form of px + q = r:
25x + 45 = 120
This equation represents the total time Jahan practiced on the weekdays (25 minutes multiplied by the number of weekdays, x) plus the time he practiced on Saturday (45 minutes) equaling the total practice time of 120 minutes.
Therefore, the equation representing the problem is 25x + 45 = 120.
On Saturday, Jahan practiced for 45 minutes.
On most weekdays, he practiced for 25 minutes each.
The total time Jahan practiced last week was 120 minutes.
To represent the problem as an equation in the form of px + q = r:
25x + 45 = 120
This equation represents the total time Jahan practiced on the weekdays (25 minutes multiplied by the number of weekdays, x) plus the time he practiced on Saturday (45 minutes) equaling the total practice time of 120 minutes.
Therefore, the equation representing the problem is 25x + 45 = 120.