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3rd monthly MATH7 nasat

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NEWFOUNDLAND
ARTS AND SCIENCE ACADEMY OF TAGUM
NUP2 Bldg., Circumferential Rd., Tagum City, Davao Del Norte
Tel. No. 216-4163 Mobile No. 0917-630-0706
THIRD MONTHLY EXAMINATION- MATH-7
Prepared by: REYMOND R. ALASAGAS
February 15-16, 2021
NAME: _____________________________________
GRADE & SECTION:__________________________
DATE: ______________
PARENT’S SIGNATURE: ___________
General Instruction:
Read the questions carefully. Follow the specified instruction below in each type of tests in your
questionnaire. Not following the given instruction will be mark WRONG.
PART1: MULTIPLE CHOICE
Choose the best and correct answer.
1. Subtract πŸπŸπ’™πŸ − πŸ”π’™π’š from πŸπŸ’π’™πŸ − πŸπŸ–π’™π’š?
a. 36π‘₯ 2 − 24π‘₯𝑦
b. 12π‘₯ 2 + 24π‘₯𝑦
c. 12π‘₯ 2 − 12π‘₯𝑦
d. 10π‘₯ 2 + 12π‘₯𝑦
2. From π’šπŸ − πŸ•π’š + 𝟏𝟐 subtract π’šπŸ -12y+28?
a. 19y-40
b. 5y-6
c. 2𝑦 2 − 19𝑦 + 16
d. 𝑦 2 − 5𝑦 − 16
3. Add πŸ”π’ŒπŸ’ to the difference of πŸ”π’ŒπŸ’ 𝒂𝒏𝒅 πŸ—π’ŒπŸ’ ?
a. 6π‘˜ 4
b. 7π‘˜ 4
c. 10π‘˜ 4
d. 16π‘˜ 4
4. What is the product of πŸ’π’‚πŸ 𝒂𝒏𝒅 πŸ“π’ƒπŸ‘ ?
a. –ab
b. 9π‘Ž2 𝑏 3
c. 20π‘Ž2 𝑏 3
d. 20ab
5. What are the two polynomials whose product is π’™πŸ + πŸ“π’™π’š + πŸ”π’šπŸ ?
a. (π‘₯ + 2𝑦)(π‘₯ + 3𝑦)
c. (π‘₯ + 5𝑦)(π‘₯ + 6𝑦)
b. π‘₯(π‘₯ + 5𝑦 + 6)
d. (π‘₯ + 2)(π‘₯ + 3)
6. What is the resulting polynomial if the sum of 3m and 5m is multiplied by the
difference of 7n and 4n?
a. 22π‘š
c. 45π‘šπ‘›
b. 24π‘šπ‘›
d. 88π‘šπ‘›
7. Find the quotient of π’™πŸ + πŸ•π’™ + 𝟏𝟐 π’ƒπ’š 𝒙 + πŸ’?
a. π‘₯ + 12
b. π‘₯ + 6
c. π‘₯ + 4
d. π‘₯ + 3
8. One of the factors of πŸ”π’‚πŸ + πŸπŸ‘π’‚ − πŸ“ π’Šπ’” πŸ‘π’‚ − 𝟏. Find the other factor?
a. 2π‘Ž + 5
c. 3π‘Ž + 5
b. 4π‘Ž − 1
d. 5π‘Ž − 1
9. What is the quotient if πŸ’π’„ − πŸ‘ is divided fromπŸ–π’„πŸ + πŸ”π’„ − πŸ—?
a. 8𝑐 + 1
c. 2𝑐 + 3
b. 4𝑐 + 2
d. 3𝑐 + 4
10. If πŸ”π’ŽπŸ‘ + πŸ•π’ŽπŸ − π’Ž + πŸ‘ is divided by πŸπ’Ž + πŸ‘, what is the other factor?
a. 6π‘š2 − 4π‘š + 3
c. 5π‘š2 − π‘š + 3
b. 3π‘š2 − π‘š + 1
d. 3π‘š2 + π‘š − 1
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