Maths-
General
Easy

Question

We can express any constant  in the variable form without changing its value as

  1. kx1
  2. kx2
  3. kx0
  4. k x to the power of 1 half end exponent

Hint:

  • A symbol which has a fixed numerical value in all cases is called constant.
  • A symbol standing in for an unknown numerical value in an equation is called variable.

The correct answer is: kx0


    • We have given a constant
    • We have to find how we can express any constant 𝑘 in the variable form without changing its value.
    Step 1 of 1:
    We have given a constant k
    We have to find how we can express any constant 𝑘 in the variable form without changing its value
    We know that x0 = 1
    So After multiplying it with any constant, it will not change its value.
    So,kx0 is the answer
    Hence, Option C is correct.

    Related Questions to study

    General
    Maths-

    x0 = ?

    x0 = ?

    Maths-General
    General
    Maths-

    State and prove the Perpendicular Bisector Theorem.

    Answer:
    • To prove: 
      • Perpendicular Bisector Theorem:

    • Proof: 
    • Statement:
      • According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
    • Step 1:
    Let consider given figure,

    Let arbitrary point C on perpendicular bisector.
    In which,
    CD is perpendicular bisector on AB.
    Hence,
    AD = DB
    CD = CD (common)
    straight angle ADC equals straight angle BDC equals 90 to the power of ring operator
    So, according to SAS rule
    straight triangle ACE approximately equal to straight triangle ACE
    Hence,
    CA = CB
    So, any point on perpendicular bisector is at equal distance from end points of line segment.
    Hence proved.

    State and prove the Perpendicular Bisector Theorem.

    Maths-General
    Answer:
    • To prove: 
      • Perpendicular Bisector Theorem:

    • Proof: 
    • Statement:
      • According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
    • Step 1:
    Let consider given figure,

    Let arbitrary point C on perpendicular bisector.
    In which,
    CD is perpendicular bisector on AB.
    Hence,
    AD = DB
    CD = CD (common)
    straight angle ADC equals straight angle BDC equals 90 to the power of ring operator
    So, according to SAS rule
    straight triangle ACE approximately equal to straight triangle ACE
    Hence,
    CA = CB
    So, any point on perpendicular bisector is at equal distance from end points of line segment.
    Hence proved.
    General
    Maths-

    What is a monomial? Explain with an example.

    Explanation:
    • We have to define monomial by giving an example.
    Step 1 of 1:
    A monomial is an expression with only one term.
    The examples are: 3x, 6y

    What is a monomial? Explain with an example.

    Maths-General
    Explanation:
    • We have to define monomial by giving an example.
    Step 1 of 1:
    A monomial is an expression with only one term.
    The examples are: 3x, 6y
    parallel
    General
    Maths-

    Find the equation of a line that passes through left parenthesis negative 3 comma 1 right parenthesis and left parenthesis 2 comma negative 14 right parenthesis

    Hint:
    We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is
    fraction numerator y minus d over denominator d minus b end fraction equals fraction numerator x minus c over denominator c minus a end fraction
    Step by step solution:
    Let the given points be denoted by
    (a, b) = (-3, 1)
    (c, d) = (2, -14)
    The equation of a line passing through two points (a, b) and (c, d) is
    fraction numerator y minus d over denominator d minus b end fraction equals fraction numerator x minus c over denominator c minus a end fraction
    Using the above points, we have
    fraction numerator y minus left parenthesis negative 14 right parenthesis over denominator negative 14 minus 1 end fraction equals fraction numerator x minus 2 over denominator 2 minus left parenthesis negative 3 right parenthesis end fraction
    Simplifying the above equation, we have
    fraction numerator y plus 14 over denominator negative 15 end fraction equals fraction numerator x minus 2 over denominator 2 plus 3 end fraction
    not stretchy rightwards double arrow fraction numerator y plus 14 over denominator negative 15 end fraction equals fraction numerator x minus 2 over denominator 5 end fraction
    Cross multiplying, we get
    5(y + 14) = -15(x - 2)
    Expanding the factors, we have
    5y + 70 = -15x + 30
    Taking all the terms in the left hand side, we have
    15x + 5y + 70 - 30 = 0
    Finally, the equation of the line is
    15x + 5y + 40 = 0
    Dividing the equation throughout by 5, we get
    3x + y + 8 = 0
    This is the required equation.
    Note:
    We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.

    Find the equation of a line that passes through left parenthesis negative 3 comma 1 right parenthesis and left parenthesis 2 comma negative 14 right parenthesis

    Maths-General
    Hint:
    We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is
    fraction numerator y minus d over denominator d minus b end fraction equals fraction numerator x minus c over denominator c minus a end fraction
    Step by step solution:
    Let the given points be denoted by
    (a, b) = (-3, 1)
    (c, d) = (2, -14)
    The equation of a line passing through two points (a, b) and (c, d) is
    fraction numerator y minus d over denominator d minus b end fraction equals fraction numerator x minus c over denominator c minus a end fraction
    Using the above points, we have
    fraction numerator y minus left parenthesis negative 14 right parenthesis over denominator negative 14 minus 1 end fraction equals fraction numerator x minus 2 over denominator 2 minus left parenthesis negative 3 right parenthesis end fraction
    Simplifying the above equation, we have
    fraction numerator y plus 14 over denominator negative 15 end fraction equals fraction numerator x minus 2 over denominator 2 plus 3 end fraction
    not stretchy rightwards double arrow fraction numerator y plus 14 over denominator negative 15 end fraction equals fraction numerator x minus 2 over denominator 5 end fraction
    Cross multiplying, we get
    5(y + 14) = -15(x - 2)
    Expanding the factors, we have
    5y + 70 = -15x + 30
    Taking all the terms in the left hand side, we have
    15x + 5y + 70 - 30 = 0
    Finally, the equation of the line is
    15x + 5y + 40 = 0
    Dividing the equation throughout by 5, we get
    3x + y + 8 = 0
    This is the required equation.
    Note:
    We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.
    General
    Maths-

    The degree of 25x2y23 is

    Explanation:
    • We have been given a polynomial expression in the question for which we have to find its degree.
    Step 1 of 1:
    We have given an expression
    We know that degree is highest power of variable present in polynomial.
    So,
    The degree of is
    Degree of  is
    Degree of is
    So, Total degree is

    Hence, Option B is correct.

    The degree of 25x2y23 is

    Maths-General
    Explanation:
    • We have been given a polynomial expression in the question for which we have to find its degree.
    Step 1 of 1:
    We have given an expression
    We know that degree is highest power of variable present in polynomial.
    So,
    The degree of is
    Degree of  is
    Degree of is
    So, Total degree is

    Hence, Option B is correct.
    General
    Maths-

    Ritu  earns R s 680 in commission and is paid R s 10.25 per hour. Karina earns R s
    410 in commissions and is paid R s 12.50 per hour. What will you find if you solve
    for x in the equation 10.25x + 680 = 12.5x + 480

    Answer:
    • Hint:
    ○   To solve the equation, group terms with the same coefficient on one side and numbers on                      the other side.
    • Step by step explanation:
    ○    Given:
    10.25x + 680 = 12.5x + 480
    ○    Step 1:
    ○    Solve the equation.
    ○    Group the like terms
    10.25x + 680 = 12.5x + 480
    rightwards double arrow680 - 480 = 12.5x - 10.25x
    rightwards double arrow200 = 2.25x
    ○    Step 2:
    ○    Divide both side with 2.25
    not stretchy rightwards double arrow fraction numerator 200 over denominator 2.25 end fraction equals fraction numerator 2.25 x over denominator 2.25 end fraction
    rightwards double arrow88.88 = x
    • Final Answer:
    Hence, the x = 88.88.

    Ritu  earns R s 680 in commission and is paid R s 10.25 per hour. Karina earns R s
    410 in commissions and is paid R s 12.50 per hour. What will you find if you solve
    for x in the equation 10.25x + 680 = 12.5x + 480

    Maths-General
    Answer:
    • Hint:
    ○   To solve the equation, group terms with the same coefficient on one side and numbers on                      the other side.
    • Step by step explanation:
    ○    Given:
    10.25x + 680 = 12.5x + 480
    ○    Step 1:
    ○    Solve the equation.
    ○    Group the like terms
    10.25x + 680 = 12.5x + 480
    rightwards double arrow680 - 480 = 12.5x - 10.25x
    rightwards double arrow200 = 2.25x
    ○    Step 2:
    ○    Divide both side with 2.25
    not stretchy rightwards double arrow fraction numerator 200 over denominator 2.25 end fraction equals fraction numerator 2.25 x over denominator 2.25 end fraction
    rightwards double arrow88.88 = x
    • Final Answer:
    Hence, the x = 88.88.
    parallel
    General
    Maths-

    If the perpendicular bisector of one side of a triangle goes through the opposite vertex, then the triangle is ____ isosceles.

    Answer:
    • Hints:
    • Perpendicular bisector theorem
    • According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
    • Step by step explanation: 
    • Step 1:
    Let consider given figure,

    Let arbitrary point C on perpendicular bisector.
    In which,
    CD is perpendicular bisector on AB.
    Hence,
    AD = DB
    CD = CD (common)
    ADC = BDC = 90o
    So, according to SAS rule
    ACD  BCD
    Hence,
    CA = CB
    So, triangle is always isosceles.
    Hence proved.
    • Final Answer: 
    Correct option A always.

    If the perpendicular bisector of one side of a triangle goes through the opposite vertex, then the triangle is ____ isosceles.

    Maths-General
    Answer:
    • Hints:
    • Perpendicular bisector theorem
    • According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
    • Step by step explanation: 
    • Step 1:
    Let consider given figure,

    Let arbitrary point C on perpendicular bisector.
    In which,
    CD is perpendicular bisector on AB.
    Hence,
    AD = DB
    CD = CD (common)
    ADC = BDC = 90o
    So, according to SAS rule
    ACD  BCD
    Hence,
    CA = CB
    So, triangle is always isosceles.
    Hence proved.
    • Final Answer: 
    Correct option A always.
    General
    Maths-

    Point P is inside △ 𝐴𝐵𝐶 and is equidistant from points A and B. On which of the following segments must P be located?

    Answer:
    • Hints:
      • Perpendicular bisector theorem
      • According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
    • Step by step explanation: 
      • Given:
    Point P is inside △ 𝐴𝐵𝐶 and is equidistant from points A and B.
    • Step 1:
    • In △ ABC,
    According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
    So,
    As P is equidistant from A and B it lies on perpendicular bisector on AB.
    • Final Answer: 
     Correct option. B. The perpendicular bisector of 𝐴𝐵.

    Point P is inside △ 𝐴𝐵𝐶 and is equidistant from points A and B. On which of the following segments must P be located?

    Maths-General
    Answer:
    • Hints:
      • Perpendicular bisector theorem
      • According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
    • Step by step explanation: 
      • Given:
    Point P is inside △ 𝐴𝐵𝐶 and is equidistant from points A and B.
    • Step 1:
    • In △ ABC,
    According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
    So,
    As P is equidistant from A and B it lies on perpendicular bisector on AB.
    • Final Answer: 
     Correct option. B. The perpendicular bisector of 𝐴𝐵.
    General
    Maths-

    A constant has a degree__________.

    Explanation:
    • We have been given a statement in the question for which we have to fill the blank by choosing the appropriate answer from the given four options.
    Step 1 of 1:
    The degree of a constant is Zero.
    By definition of degree we know that it is highest power of variable present in polynomial.
    But for constant there is no polynomial.
    So, The degree of constant is Zero
    Hence, Option B is correct.

    A constant has a degree__________.

    Maths-General
    Explanation:
    • We have been given a statement in the question for which we have to fill the blank by choosing the appropriate answer from the given four options.
    Step 1 of 1:
    The degree of a constant is Zero.
    By definition of degree we know that it is highest power of variable present in polynomial.
    But for constant there is no polynomial.
    So, The degree of constant is Zero
    Hence, Option B is correct.
    parallel
    General
    Maths-

    Graph the equation straight Y equals 3 straight X minus 5

    Hint:
    To plot the graph of an equation, first we make a table of points satisfying that equation. Then we draw an x-axis and y-axis on the graph. After that we scale both the axis according the values we get in the table. Lastly, we plot the points from the table on the graph and join them to get the required curve.
    Step by step solution:
    The given equation is
    y = 3x − 5
    First we make a table of points satisfying the equation.
    Putting x = 0 in the above equation, we will get, y = -5
    Similarly, putting x = 1, in the above equation, we get y = −2
    Continuing this way, we have
    For x = -1, we get y = -8
    For x = 2, we get y = 1
    For x = -2, we get y = -11
    Making a table of all these points, we have

    Now we plot these points on the graph.

    After plotting the points, we join them with a line to get the graph of the equation.
    Note:
    We can find the tabular values for any points of x and then plot them on the graph. But we usually choose values for which calculating y is easier. This makes plotting the graph simpler. We can also find the values by putting different values of y in the equation to get different values for x. Either way, we need points satisfying the equation to plot its graph

    Graph the equation straight Y equals 3 straight X minus 5

    Maths-General
    Hint:
    To plot the graph of an equation, first we make a table of points satisfying that equation. Then we draw an x-axis and y-axis on the graph. After that we scale both the axis according the values we get in the table. Lastly, we plot the points from the table on the graph and join them to get the required curve.
    Step by step solution:
    The given equation is
    y = 3x − 5
    First we make a table of points satisfying the equation.
    Putting x = 0 in the above equation, we will get, y = -5
    Similarly, putting x = 1, in the above equation, we get y = −2
    Continuing this way, we have
    For x = -1, we get y = -8
    For x = 2, we get y = 1
    For x = -2, we get y = -11
    Making a table of all these points, we have

    Now we plot these points on the graph.

    After plotting the points, we join them with a line to get the graph of the equation.
    Note:
    We can find the tabular values for any points of x and then plot them on the graph. But we usually choose values for which calculating y is easier. This makes plotting the graph simpler. We can also find the values by putting different values of y in the equation to get different values for x. Either way, we need points satisfying the equation to plot its graph
    General
    Maths-

    A= 5x-3y +2z, B= 4x-2y+3z, C= 6x-4y -4z, Find A-B+C

    Answer:
    • Hint:
    ○    Put values of A, B, and C in expression A-B+C to get the answer.
    • Step by step explanation:
    ○    Given:
    A= 5x-3y +2z,
    B= 4x-2y+3z,
    C= 6x-4y -4z,
    Find A-B+C
    ○    Step 1:
    ○    Put values of A, B and C in (A-B+C)
    We get,
    A-B+C
    rightwards double arrow (5x-3y +2z) - (4x-2y+3z) + (6x-4y -4z)
    ○    Step 2:
    ○    Group like terms
    rightwards double arrow (5x-4x +6x) - (3y-2y+4y) + (2z-3z -4z)
    rightwards double arrow 7x - 5y - 5z
    • Final Answer:
    Hence, the value of A-B+C is 7x - 5y - 5z.

    A= 5x-3y +2z, B= 4x-2y+3z, C= 6x-4y -4z, Find A-B+C

    Maths-General
    Answer:
    • Hint:
    ○    Put values of A, B, and C in expression A-B+C to get the answer.
    • Step by step explanation:
    ○    Given:
    A= 5x-3y +2z,
    B= 4x-2y+3z,
    C= 6x-4y -4z,
    Find A-B+C
    ○    Step 1:
    ○    Put values of A, B and C in (A-B+C)
    We get,
    A-B+C
    rightwards double arrow (5x-3y +2z) - (4x-2y+3z) + (6x-4y -4z)
    ○    Step 2:
    ○    Group like terms
    rightwards double arrow (5x-4x +6x) - (3y-2y+4y) + (2z-3z -4z)
    rightwards double arrow 7x - 5y - 5z
    • Final Answer:
    Hence, the value of A-B+C is 7x - 5y - 5z.
    General
    Maths-

    For an obtuse triangle, the circumcentre lies

    Answer:
    For obtuse triangle, the circumcentre lies outside the triangle.

    For an obtuse triangle, the circumcentre lies

    Maths-General
    Answer:
    For obtuse triangle, the circumcentre lies outside the triangle.
    parallel
    General
    Maths-

    For a right triangle, the circumcentre lies

    Answer:
    For a right triangle, the circumcentre lies on the triangle.
    a. on the triangle.
     

    For a right triangle, the circumcentre lies

    Maths-General
    Answer:
    For a right triangle, the circumcentre lies on the triangle.
    a. on the triangle.
     
    General
    Maths-

    How does adding and subtracting polynomials compare to adding and subtracting integers?

    Explanation:
    • We have to find out how does adding and subtracting polynomials compare to adding and subtracting integers.
    Step 1 of 1:
    In adding and subtracting of integers, only numbers are involved but in adding and subtracting polynomial there is involvement of variable and constants both.
    In polynomial, we can add only like terms.

    How does adding and subtracting polynomials compare to adding and subtracting integers?

    Maths-General
    Explanation:
    • We have to find out how does adding and subtracting polynomials compare to adding and subtracting integers.
    Step 1 of 1:
    In adding and subtracting of integers, only numbers are involved but in adding and subtracting polynomial there is involvement of variable and constants both.
    In polynomial, we can add only like terms.
    General
    Maths-

    For an acute triangle, the circumcentre lies

    Answer:
    For an acute triangle, the circumcentre lies inside the triangle.
    b. inside the triangle.

    For an acute triangle, the circumcentre lies

    Maths-General
    Answer:
    For an acute triangle, the circumcentre lies inside the triangle.
    b. inside the triangle.
    parallel

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