Introduction for Solving Mathematics Grind:
Well experienced tutor (or) graduate Mathematics student offering Junior and Leaving Certificate grinds around Ireland. Grinds contains various subjects such as Mathematics, French, Accounting etc., in this math grind covers all level of students. They are providing explanation of theory with detailed examples with questions and answers. In this article we shall discuss about solving mathematics grind. The following examples are involved in solving mathematics grind. Is this topic Exponential Function Solver hard for you? Watch out for my coming posts.
Solving Mathematics Grind Example: 1
Solve the sum and find the value of ‘x’
50x + 40 = -200
Subtract 40 from both sides:
50x + 40 – 40 = -200 - 40
Simplify both sides:
50x = -240
Divide both sides by 50:
`(50x)/50` = `-240/50`
Simplify both sides:
x = `-24/5`
I have recently faced lot of problem while learning Place Value with Decimals, But thank to online resources of math which helped me to learn myself easily on net.
Solving Mathematics Grind Example: 2
Use Euclid’s algorithms solve the HCF of 2245 and 36548.
Solution:
Since 36548 > 2245, we apply the division lemma to 36548 and 2245, to get
36548 = 2245 × 16 + 628
Since the remainder 628 not equal to 0, we apply the division lemma to 2245 and 628, to get
2245 = 628 × 3 + 361
We consider the new divisor 628 and the new remainder 361, and apply the division lemma to get
628 = 361 × 1 + 267
We consider the new divisor 361 and the new remainder 267, and apply the division lemma to get
361 = 267 × 1 + 94
We consider the new divisor 267 and the new remainder 94, and apply the division lemma to get
267 = 94 × 2 + 79
We consider the new divisor 94 and the new remainder 79, and apply the division lemma to get
94 = 79 × 1 + 15
We consider the new divisor 79 and the new remainder 15, and apply the division lemma to get
79 = 15 × 5 + 4
We consider the new divisor 15 and the new remainder 4, and apply the division lemma to get
15 = 4 × 3 + 3
We consider the new divisor 4 and the new remainder 3, and apply the division lemma to get
4 = 3 × 1 + 1
We consider the new divisor 3 and the new remainder 1, and apply the division lemma to get
3 = 1 × 3 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 1, the HCF of 36548 and 2245 is 1.
Notice that 1 = HCF (3, 1) = HCF (4, 3) = HCF (15, 4) = HCF (79, 15) = HCF (94, 79) = HCF (267, 94) = HCF (361, 267) = HCF (628, 361) = HCF (2245, 628) = HCF (36548, 2245).
Well experienced tutor (or) graduate Mathematics student offering Junior and Leaving Certificate grinds around Ireland. Grinds contains various subjects such as Mathematics, French, Accounting etc., in this math grind covers all level of students. They are providing explanation of theory with detailed examples with questions and answers. In this article we shall discuss about solving mathematics grind. The following examples are involved in solving mathematics grind. Is this topic Exponential Function Solver hard for you? Watch out for my coming posts.
Solving Mathematics Grind Example: 1
Solve the sum and find the value of ‘x’
50x + 40 = -200
Subtract 40 from both sides:
50x + 40 – 40 = -200 - 40
Simplify both sides:
50x = -240
Divide both sides by 50:
`(50x)/50` = `-240/50`
Simplify both sides:
x = `-24/5`
I have recently faced lot of problem while learning Place Value with Decimals, But thank to online resources of math which helped me to learn myself easily on net.
Solving Mathematics Grind Example: 2
Use Euclid’s algorithms solve the HCF of 2245 and 36548.
Solution:
Since 36548 > 2245, we apply the division lemma to 36548 and 2245, to get
36548 = 2245 × 16 + 628
Since the remainder 628 not equal to 0, we apply the division lemma to 2245 and 628, to get
2245 = 628 × 3 + 361
We consider the new divisor 628 and the new remainder 361, and apply the division lemma to get
628 = 361 × 1 + 267
We consider the new divisor 361 and the new remainder 267, and apply the division lemma to get
361 = 267 × 1 + 94
We consider the new divisor 267 and the new remainder 94, and apply the division lemma to get
267 = 94 × 2 + 79
We consider the new divisor 94 and the new remainder 79, and apply the division lemma to get
94 = 79 × 1 + 15
We consider the new divisor 79 and the new remainder 15, and apply the division lemma to get
79 = 15 × 5 + 4
We consider the new divisor 15 and the new remainder 4, and apply the division lemma to get
15 = 4 × 3 + 3
We consider the new divisor 4 and the new remainder 3, and apply the division lemma to get
4 = 3 × 1 + 1
We consider the new divisor 3 and the new remainder 1, and apply the division lemma to get
3 = 1 × 3 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 1, the HCF of 36548 and 2245 is 1.
Notice that 1 = HCF (3, 1) = HCF (4, 3) = HCF (15, 4) = HCF (79, 15) = HCF (94, 79) = HCF (267, 94) = HCF (361, 267) = HCF (628, 361) = HCF (2245, 628) = HCF (36548, 2245).
