The value of x is 75.
what is Algebra?A branch of mathematics known as algebra deals with symbols and the mathematical operations performed on them.
Variables are the name given to these symbols because they lack set values.
In order to determine the values, these symbols are also subjected to various addition, subtraction, multiplication, and division arithmetic operations.
Given:
6 consecutive integers are x − 2, x − 1, x, x + 1, x + 2, x + 3
Now, the integers add up to 447.
So, x-2 + x-1 + x+ x +1 + x+2 + x+3= 447
6x -3 + 6 = 447
6x = 450
x= 75
Hence, the value of x is 75.
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Give one example of an equation with variables on both sides that has all real numbers as the solution. Also give an example of an equation with variables on both sides that has no real solutions.
An equation with all real numbers as solutions is one that, regardless of what value a variable might have, both sides will always be equal, such as 2x+3 = 2x+3. An equation with no real solutions is a statement where the two sides cannot ever be equal, like 2x+5 = 2x+7.
Explanation:An example of an equation with variables on both sides that has all real numbers as the solution is a statement where both sides are identical. For instance, 2x+3 = 2x+3. All real numbers can satisfy this equation for x because both sides are always equal, regardless of what your value of x is.
On the other hand, an equation with no real solutions is one where it's impossible to find any real number that satisfies the equation. An example can be 2x+5 = 2x+7. No matter what value you put in place of x, the two sides of the equation will never be equal. Hence, this equation has no real solution.
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How do you find the measures of the angles in a rhombus with the given angle 54 degrees?
If m is the midpoint of pq, what is the relationship between pm and pq
Midpoint, as the word suggests, means the point which lies in the middle of something. If m is the midpoint of PQ then PM will be half of PQ.
What does a midpoint mean?Midpoint, as the word suggests, means the point which lies in the middle of something. The midpoint of a line segment means a point that lies in the mid of the given line segment.
Given that m is the midpoint of PQ. Therefore, it will divide PQ into two equal parts. Therefore, we can write,
PQ = PM + MQ
But since PM = MQ,
PQ = PM + PM
PQ = 2 PM
PM = PQ/2
Hence, if m is the midpoint of PQ then the relationship between PM and PQ will be half of PQ.
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Eight times a number is not less than the sum of the number and fifty
What is 7 1/5 written as a percent
2,438,783 divided 893
Answer:
quotient=2,731
Remainder=0
Step-by-step explanation:
893) 2,438,783 (2,731
- 1 786
---------
6527
- 6251
---------
2768
- 2679
-----------
893
- 893
---------
0
Solve for x. 45(15x+20)−7x=56(12x−24)+6
Equations may be true or false, just as word sentences may be true or false. The equation:
3 + x = 7
will be false if any number except 4 is substituted for the variable. The value of the variable for which the equation is true (4 in this example) is called the solution of the equation. We can determine whether or not a given number is a solution of a given equation by substituting the number in place of the variable and determining the truth or falsity of the result.
Example 1 Determine if the value 3 is a solution of the equation
4x - 2 = 3x + 1
Solution We substitute the value 3 for x in the equation and see if the left-hand member equals the right-hand member.
4(3) - 2 = 3(3) + 1
12 - 2 = 9 + 1
10 = 10
Ans. 3 is a solution.
The first-degree equations that we consider in this chapter have at most one solution. The solutions to many such equations can be determined by inspection.
Example 2 Find the solution of each equation by inspection.
a. x + 5 = 12
b. 4 · x = -20
Solutions a. 7 is the solution since 7 + 5 = 12.
b. -5 is the solution since 4(-5) = -20.
Find the greatest common factor 46m4 22m3
Identify the fraction as proper, improper, or mixed . (If the fraction is equivalent to 1, choose "one" in the blank.)
2.4 puzzle time how can you share five apples with seven friends
1. The algebraic property of equality that is represented is the multiplication property. The answer is letter P.
2. The algebraic property of equality that is represented is the subtraction
property. The answer is letter E.
3. The algebraic property of equality that is represented is the substitution
property. The answer is letter M.
4. The algebraic property of equality that is represented is the division
property. The answer is letter A.
5. The algebraic property of equality that is represented is the addition
property. The answer is letter A.
6. The algebraic property of equality that is represented is the distributive
property. The answer is Letter E.
7. The algebraic property of equality that are represented is the transitive
property. The answer is letter is E.
8. The algebraic property of equality that are represented is the symmetric
property. The answer is Letter A.
9. The algebraic property of equality that are represented is the reflexive
property. The answer is letter C.
10. The algebraic property of equality that are represented by is the multiplication property.
The answer is letter P.
11. The algebraic property of equality that are represented is the subtraction property.
The answer is letter L.
12. The algebraic property of equality that are represented is the reflexive property.
The answer is letter S.
13. The algebraic property of equality that are represented is the transitive property.
The answer is letter U.
14. The algebraic property of equality that are represented is the symmetric property.
The answer is letter K.
The given is 3 5 14 2 4 1 10 11 6 12 8 13 9 7
Plugging in the letters we got for each number, gives us MAKE APPLESAUCE.
As the number of apples is less than the total number of friends among which the apples need to be distributed so, no one will get a complete piece of apple so, 5 apples can be shared in between 7 friends as [tex]\frac{5}{7}[/tex].
According to the question, the total number of apples are 5 and the total number of friends are 7.
So, each friend will get (bu unitary method)-
[tex]\rm{Each\;friend\;will\;get}=\dfrac{5}{7}\;\rm apples[/tex]
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what is 100/6 as a decimal
Ali currently has $25. He is going to start saving $5 every week.
Which equation represents this situation?
y = 25x – 5
y = 5x + 25
y = 25x + 5
y = 5x – 25
Problem page the perimeter of a rectangular garden is 274 feet. if the width of the garden is 63 feet, what is its length?
How can you find the measure of a distance that you cannot measure directly?
12 foot by 15 foot swimming pool has a 3 ft wide no slip surface around it what is the outer perimeter of the no slip surface
the pool is 12 x 15
add 6ft to each dimension ( 3 feet on each side)
12+6 = 18
15+6 = 21
21 *2 = 42
18*2 = 36
42 +36 = 78 feet perimeter
How to solve 16x to the 4th power y to the 3rd power divided by 2x to the 4th power?
(3^-2)^4 *3^3 divided by 3^-3
You are making identical gift bags using 24 candles and 36 bottles of lotion .what is the greatest number of gift bags you can make with no items left over. Show your work
Answer: 12
Step-by-step explanation:
Given : You are making identical gift bags using 24 candles and 36 bottles of lotion .
Then, the greatest number of gift bags you can make with no items left over= greatest common factor of 24 and 36
Prime factorization of 24 and 36 :_
[tex]24=2\times2\times2\times3\\\\ 36=2\times2\times3\times3[/tex]
GCF(24 , 36)= [tex]2\times2\times3=12[/tex]
Hence, the greatest number of gift bags you can make with no items left over =12
Enter an inequality that represents the graph in the box.
PLEASE HELP ME!
its actually y=>3x-1 because it's shaded below the line.
below the line: < or <=
above the line: > or>=
Three friends rented a kayak. It cost $4 per hour per person to rent the kayak, plus $2 for each life jacket, and $3 to park the car. It cost $57 in all. How many hours did they spend kayaking? Write an equation and solve.
The circumference of a circle is 25.12 yards. What is the radius of the circle?
Write an equation that shows the equivalency between meters and gigameters.
To convert between meters and gigameters, use the equation 1 Gm = 10^9 m. This relationship is important for converting units in scientific and mathematical contexts.
Explanation:To show the equivalency between meters and gigameters, we can use the following equation:
1 gigameter (Gm) = 10⁹ meters (m).
This equation means that one gigameter is equal to one billion meters. We can use this equivalency for converting measurements from meters to gigameters or vice versa. For example, to convert 2 gigameters to meters, simply multiply 2 by 10⁹ to get 2 x 10⁹ meters.
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If ab= 37 and xy =1/37, what is the value of the product x x b x y x a
In this case, we are simply asked to find for the value of:
xbya or abxy
Since we know that given values of ab and xy, therefore abxy is:
abxy = 37 * (1 / 37)
abxy = 37 / 37
abxy = 1
3/5y=-7/25 solve for y
given the functon g(x)= (1/4)^x, determine the y-intercept and the equation of the asymptote.
a. the y-intercept is (0, 0.25) and the asymptote is at y=0
b. the y-intercept is (1,0) and the asymptote is at y=0. the y-intercept is (1,0) and the asymptote is at y=1
c. the y-intercept is (0,1) and the asymptote is at y=0
d. the y-intercept is (1,1) and the asymptote is at y= 1.
The y-intercept of the function g(x) = (1/4)^x is (0,1), and the horizontal asymptote is at y=0.
The correct option is (c).
Given the function g(x) = (1/4)^x, we need to find the y-intercept and the equation of the horizontal asymptote.
The y-intercept occurs where x = 0. Substituting 0 for x in g(x), we get:
g(0) = (1/4)^0 = 1
So the y-intercept is at the point (0,1).
For horizontal asymptotes, we look at the behavior of the function as x goes to positive or negative infinity. In this case, as x goes to infinity, (1/4)^x approaches 0. Therefore, the horizontal asymptote of g(x) is at y = 0.
The correct answer is c. The y-intercept is (0,1) and the asymptote is at y=0.
64 ounces in 8 cups unit rate
Solve for x. x - 3/8 = - 5/8
x = 1/4
x = - 1/4
x = -1
x = 1
Answer:
-1/4 should be the answer
Step-by-step explanation:
An art store offers prints in two sizes. The store earns $15 on each small print sold and $25 on each large print sold. The owner needs to make a daily profit of at least $700 in order to cover costs. Due to equipment limitations, the number of small prints made must be more than three times the number of large prints. Given that x represents the number of small prints sold and y represents the number of large prints sold, determine which inequalities represent the constraints for this situation. x + y ≤ 60 15x + 25y < 700 x > 3y 15x + 25y ≥ 700 y > 3x x + 3y ≥ 60 Which combinations of small prints and large prints satisfy this system? (45,10) (35,15) (30,10) (40,5)
The constraints for this problem are represented by the inequalities x > 3y and 15x + 25y ≥ 700. The combinations that satisfy the constraints is (45,10) as it meets both conditions.
Explanation:The constraints for the art store situation are represented by the inequalities x > 3y and 15x + 25y ≥ 700. The first inequality, x > 3y, represents the condition that the number of small prints made must be more than three times the number of large prints sold. The second inequality, 15x + 25y ≥ 700, characterizes the fact the store needs to make a daily profit of at least $700 in order to cover costs.
With these inequalities, we can determine which pairs of small print and large print quantities satisfy this system of constraints. Checking each pair, we find that the combination (45,10) satisfies both inequalities as 45 is more than three times 10, and 15*45 + 25*10 = $775, which is greater than $700. Thus, selling 45 small prints and 10 large prints will satisfy the constraints.
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Which is farther 200km or 200mi
Answer:
200 mi.
Step-by-step explanation:
200 mi. is more or greater than 200 km. Because, 200 mi. in km will be 321 km so that is greater than 200 km.
I hope this helps you.
If 3,200 cars were sold all year, how many cars were sold in the jan/feb period?
12 months per year
3200/12 = 266.66 per month
jan/feb = 2 months
266.666 x 2 = 533.33 round off to 533 cars sold
If 3,200 cars were sold all year, then approximately 533 cars were sold in the Jan/Feb period.
We need to calculate the proportion of the year that these two months represent and then apply that proportion to the total number of cars sold in the year.
Since there are 12 months in a year, the Jan/Feb period represents [tex]\(\frac{2}{12}\)[/tex] of the year. We can simplify this fraction to [tex]\(\frac{1}{6}\)[/tex] by dividing both the numerator and the denominator by 2.
To find the number of cars sold in the Jan/Feb period, we multiply the total number of cars sold in the year by the fraction representing Jan/Feb:
[tex]\[ \text{Number of cars sold in Jan/Feb} = \text{Total cars sold} \times \frac{2}{12} \][/tex]
Substituting the given total number of cars sold:
[tex]\[ \text{Number of cars sold in Jan/Feb} = 3,200 \times \frac{2}{12} \][/tex]
[tex]\[ \text{Number of cars sold in Jan/Feb} = 3,200 \times \frac{1}{6} \][/tex]
[tex]\[ \text{Number of cars sold in Jan/Feb} = \frac{3,200}{6} \][/tex]
Performing the division:
[tex]\[ \text{Number of cars sold in Jan/Feb} = 533.\overline{3} \][/tex]
[tex]\[ \text{Number of cars sold in Jan/Feb} \approx 533[/tex]