Factorise 3a squared + 7a
Answers
Answer 1
Answer:
GCF: 1 (3 cant go into 7 and the other way around in whole number terms.)
So, 1 is the factor you are distributing.
1 goes into 3 three times.
1 goes into 7 seven times.
Don't forget your variable!
So the answer is: a(3a+ 7).
So, 1 is the factor you are distributing.
1 goes into 3 three times.
1 goes into 7 seven times.
Don't forget your variable!
So the answer is: a(3a+ 7).
Answer 2
Answer:
to factorise this we have to first find the biggest common factor. 3 and 7 are both are prime numbers so we can't factorise it by a number. but both numbers have an 'a' so we have to factorise it by a. hence the answer is,
a(3a+7).
hope it helps.
a(3a+7).
hope it helps.
Related Questions
A college graduate expects to earn a salary of $55,000 during the first year after graduation and receive a 4% raise every year after that. What is the total income he will have received after ten years?
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4.5.11Mohamed wrote the paragraph proof to show that ADEG AEFG. What mistake did he make?ADEG and AEFG are right triangles.The figure shows DE = EF, ANDEG EG by the Reflexive Property.Therefore, by the HL theorem,ADEG AEFGGX ХChoose the correct answer below.O A. The two triangles are congruent by SAS, not the HL Theorem.O B. In AEFG, EF is a leg and in ADEG, DE is the hypotenuse, so they cannot be corresponding sides.O C. The two triangles are congruent by SSS, not the HL Theorem.OD. In AEFG, EG is a leg, and in ADEG, EG is the hypotenuse, so they cannot be corresponding sides.
The quotient of 5/31 divided by 15/23, reduced to the lowest fraction, is A. 115/465. B. 23/93. C. 75/373. D. 93/23 or 4 1/23.
Round 34 449 to the a) nearest 10. b) nearest 100. c) nearest 1000.
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4.5.11Mohamed wrote the paragraph proof to show that ADEG AEFG. What mistake did he make?ADEG and AEFG are right triangles.The figure shows DE = EF, ANDEG EG by the Reflexive Property.Therefore, by the HL theorem,ADEG AEFGGX ХChoose the correct answer below.O A. The two triangles are congruent by SAS, not the HL Theorem.O B. In AEFG, EF is a leg and in ADEG, DE is the hypotenuse, so they cannot be corresponding sides.O C. The two triangles are congruent by SSS, not the HL Theorem.OD. In AEFG, EG is a leg, and in ADEG, EG is the hypotenuse, so they cannot be corresponding sides.
The quotient of 5/31 divided by 15/23, reduced to the lowest fraction, is A. 115/465. B. 23/93. C. 75/373. D. 93/23 or 4 1/23.
Round 34 449 to the a) nearest 10. b) nearest 100. c) nearest 1000.
Answers
Car see it theninlanfe
Answer: 7
Step-by-step explanation:
B. 2u - 6 R 11
C. 2u + 6 R 11
Answers
(2u² - 7) ÷ (u + 3)
2u² + 0u - 7 ÷ u + 3
2u - 6
u + 3 I 2u² + 0u - 7
- 2u² + 6u
0 - 6u - 7
+ - 6u - 18
0 + 11
Answer is: B. 2u - 6 R 11
2u² + 0u - 7 ÷ u + 3
2u - 6
u + 3 I 2u² + 0u - 7
- 2u² + 6u
0 - 6u - 7
+ - 6u - 18
0 + 11
Answer is: B. 2u - 6 R 11
Answers
Answer:
b7 or b
5
Step-by-step explanation:
6b-141 or 2b+1
11
+1 to both sides to get 6b42
then divide both sides by 6 to get b7
then for the other side you -1 by both sides to get 2b10
then divide both sides by 2 to get b5
then your final answer would be b7 or b
5.