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9th Grade Math Problems with Answers
Free 9th grade math problem with answer are provided for students practice of the subject. The students get an idea or an overview on the topic before attending the exam. Work solution is provided on different topic of class 9th like rationalize, percentage, solving the equation, graph,etc.
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We have given a set of 9th grade maths problem with answer.
1. Rationalism the denominator:
$\frac{2\sqrt{3} + 3\sqrt{3}}{2\sqrt{3}  3\sqrt{2}}$
Solution
= $\frac{2\sqrt{3} + 3\sqrt{3}}{2\sqrt{3}  3\sqrt{2}}$ * $\frac{2\sqrt{3} + 3\sqrt{2}}{2\sqrt{3} + 3\sqrt{2}}$
= $\frac{4*3+9*2+2*2\sqrt{3}*3\sqrt{2}}{(2\sqrt{3})^2(3\sqrt{2})^2}$
= $\frac{12+18+12\sqrt{6}}{1218}$
= $\frac{30+12\sqrt{6}}{6}$
2. The sum of the squares of the two larger of three consecutive even integers is 14 less than 2 times the square of the smaller one. Find the even numbers.
Solution: 
Let the numbers be x,x – 2 , x and x + 2.
Given the sum of the square of the integers is 14 less than 2 times the square of the smaller one.
That is x^{2}+ (x + 2)^{2 }= 2(x  2)^{2 } 14
x^{2 }+ x^{2} + 4x + 4 = 2(x^{2 } 4x + 4  14
2x^{2} + 4x + 4 = 2x^{2 } 8x + 8  14 subtract 2x2 on both side
4x + 4 = 8x  6 add 4x on both side
4x + 4  4x = 8x 6  4x
4 = 12x  6 subtracting 4 on both side
0 = 12x  10 add 10 on both side
10 = 12x divide by constant term 12 on both side
$\frac{10}{12}= x$
$\frac{5}{6}= x$
3. Solve the linear equation to find the value of x
$\frac{5x}{4}$ + 3x = 18
Solution:
Step 1: take the LCM and simplify LHS
$\frac{10}{12}= x$
$\frac{5}{6}= x$
3. Solve the linear equation to find the value of x
$\frac{5x}{4}$ + 3x = 18
Solution:
Step 1: take the LCM and simplify LHS
$\frac{5x+12x}{4}$ = 18
$\frac{17x}{4}$ = 18
Step 2: multiply by a constant term $\frac{4}{17}$ on both side to simplify the equation.
$\frac{17x}{4}*\frac{4}{17}$ = 18 * $\frac{4}{17}$
x = $\frac{18 * 4}{17}$
x = $\frac{72}{17}$
4. What is the standard form of equation
The standard form of equation is ax^{2 }+ bx + c = 0
5. Find the circumference of a circular disk whose area is 100pi square centimeters.
Solution: Given area of a circular disk = 100πc^{2} Area of circle = πr^{2} assign the value of area in the equation
5. Find the circumference of a circular disk whose area is 100pi square centimeters.
Solution: Given area of a circular disk = 100πc^{2} Area of circle = πr^{2} assign the value of area in the equation
100π = πr^{2} divide by a constant term π on both side
100= r^{2} taking square root on both side of equation
√100 = √ r^{2}
10 = r
the radius of circular disk = 10cm
we have the formula for circumference of circle, C = 2 πr
C = 2* π*10 assign the value of π = 22/7
= 62.89
circumference of a circular disk = 62.89 cm
6. Find an equation of the line containing ( 4,5) and perpendicular to the line 5x  3y = 4.
circumference of a circular disk = 62.89 cm
6. Find an equation of the line containing ( 4,5) and perpendicular to the line 5x  3y = 4.
Solution: Given perpendicular to the line 5x  3y = 4.
writing the equation in the form of slope of the equation y = m x+b,
5x  3y = 4. subtract by 5x on both side
 3y = 12  5x divide by constant number 3 on both side
 y = 4  $\frac{5}{3}$ x taking – as common
 y = 4  $\frac{5}{3}$ x taking – as common
y = $\frac{5}{3}$ x 4
the equation in the form of y = mx+b
the slope of equation = (m ) = $\frac{5}{3}$
the yintercept = (b) =  4
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Question:What Is the answer to these 9th grade math problems?
1. 1(a  10) < 10(a  4) + 3a
2. 3 + 11(12 + 9) < 6(n + 6)
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The answer will end up looking like this a < 7 and for number 2 its grater than or equal to.
Answers:Easy problems. 1. Solve like an equation. 1(a10) < 10(a4) + 3a, simplify both sides of the equation a+10 < 10a+40 + 3a a+10 < 7a+40 Now add a to both sides. 10 < 6a +40, subtract 40 from the right side. 30 < 6a, divide both sides by 6 (when you divide by negative number, the sign changes) 5>a. The answer is a < 5. 2. 3 + 11(12+9) <6(n+6), simplify both sides again. 3 + 132+99 < 6n + 36 228 < 6n + 36, subtract 36 from both sides. 192 < 6n, divide by 6 on both sides. 32 < n n > 32 is the answer.
Answers:Easy problems. 1. Solve like an equation. 1(a10) < 10(a4) + 3a, simplify both sides of the equation a+10 < 10a+40 + 3a a+10 < 7a+40 Now add a to both sides. 10 < 6a +40, subtract 40 from the right side. 30 < 6a, divide both sides by 6 (when you divide by negative number, the sign changes) 5>a. The answer is a < 5. 2. 3 + 11(12+9) <6(n+6), simplify both sides again. 3 + 132+99 < 6n + 36 228 < 6n + 36, subtract 36 from both sides. 192 < 6n, divide by 6 on both sides. 32 < n n > 32 is the answer.
Question:Ok, so all you need to know is that I'm going to take the SHSAT soon, and I need more help. I want a question(s) that is/are based on only 6th grade knowledge, but needs a lot of thought and effort to solve.
And can you give me an explanation on how to get the answer please ? THank you soo much !
Answers:Hello! Find the average of the sum of the numbers 1  1000. Explanation: 1 + 2 + 3 ... + 998 + 999 + 1000 Here we see if we add the first and last numbers, we always get 1001: (1000 + 1 = 1001, 999 + 2 = 1001) We do this 500 times, since we take two integers at a time, Thus: 1001 * 500 ==> 500500 Dividing that by 1000, we get our final answer: = 500.5 I hope this helps! Sincerely, Mr.Math
Answers:Hello! Find the average of the sum of the numbers 1  1000. Explanation: 1 + 2 + 3 ... + 998 + 999 + 1000 Here we see if we add the first and last numbers, we always get 1001: (1000 + 1 = 1001, 999 + 2 = 1001) We do this 500 times, since we take two integers at a time, Thus: 1001 * 500 ==> 500500 Dividing that by 1000, we get our final answer: = 500.5 I hope this helps! Sincerely, Mr.Math
Question:ok so i learned last year how to solve this kind of problems but now i dont remember please help mee... here s one example: 7x + 4x = 33 please solve it and show procedure and if you can explain what you did ooh and solve for x thanks and please do it as fast as possible
Answers:7x+4x=33 11x=33 x=3
Answers:7x+4x=33 11x=33 x=3