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kkurt [141]
3 years ago
8

If a polygon has four right angles, then it is either a square or a rectangle.

Mathematics
1 answer:
morpeh [17]3 years ago
5 0
B. A polygon is either a square or a rectangle
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Pls help i will give you a 5 star rating
docker41 [41]

                                           Question # 1

Given the expression

6^2\div \:3+\left(5+3\cdot \:2\right)-2^3

Follow the PEMDAS order of operations

\mathrm{Calculate\:within\:parentheses}\:\left(5+3\cdot \:2\right)\::\quad 11

=6^2\div \:3+11-2^3

\mathrm{Calculate\:exponents}\:6^2\::\quad 36

=36\div \:3+11-2^3

\mathrm{Calculate\:exponents}\:2^3\::\quad 8

=36\div \:3+11-8

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:36\div \:3\::\quad 12

=12+11-8

\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:12+11-8\::\quad 15

=15

Therefore,

6^2\div \:3+\left(5+3\cdot \:2\right)-2^3=15

                                            Question # 2

Given the expression

\frac{4+3^2-15\div 5}{\left(2^4-5\cdot \:3\right)^2}

as

4+3^2-\frac{15}{5}

=4+9-\frac{15}{5}      ∵3^2=9

\mathrm{Add\:the\:numbers:}\:4+9=13

=-\frac{15}{5}+13

and

\left(2^4-5\cdot \:3\right)^2

=\left(16-5\cdot \:3\right)^2   ∵ 2^4=16

\mathrm{Multiply\:the\:numbers:}\:5\cdot \:3=15

=\left(16-15\right)^2

\mathrm{Subtract\:the\:numbers:}\:16-15=1

=1^2

\mathrm{Apply\:rule}\:1^a=1

=1

Thus the equation \frac{4+3^2-15\div 5}{\left(2^4-5\cdot \:3\right)^2}  becomes

=\frac{-\frac{15}{5}+13}{1}

\mathrm{Divide\:the\:numbers:}\:\frac{15}{5}=3

=\frac{-3+13}{1}

\mathrm{Apply\:rule}\:\frac{a}{1}=a

=-3+13

\mathrm{Add/Subtract\:the\:numbers:}\:-3+13=10

=10

Therefore,

\frac{4+3^2-\frac{15}{5}}{\left(2^4-5\cdot \:3\right)^2}=10

                                                       Question # 3

Given the expression

ab-c^2+2b

Putting a = 2, b = 4, and c = 1 in the expression

=\left(2\right)\left(4\right)-\left(1\right)^2+2\left(4\right)

Follow the PEMDAS order of operations

\mathrm{Calculate\:exponents}\:\left(1\right)^2\::\quad 1

=\left(2\right)\left(4\right)-1+2\left(4\right)

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:\left(2\right)\left(4\right)\::\quad 8

=8-1+2\left(4\right)

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:2\left(4\right)\::\quad 8

=8-1+8

\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:8-1+8\::\quad 15

=15

Therefore,

ab-c^2+2b=\left(2\right)\left(4\right)-\left(1\right)^2+2\left(4\right)=15

                                                     Question # 4

Given the expression

4d^3+2e\div \:f+de

Putting d = 2, e = 3, and f = 6 in the expression

=4\left(2\right)^3+2\left(3\right)\div \:6+\left(2\right)\left(3\right)

Follow the PEMDAS order of operations

\mathrm{Calculate\:exponents}\:\left(2\right)^3\::\quad 8

=4\cdot \:8+2\left(3\right)\div \:6+\left(2\right)\left(3\right)

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:4\cdot \:8\::\quad 32

=32+2\left(3\right)\div \:6+\left(2\right)\left(3\right)

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:2\left(3\right)\div \:6\::\quad 1

=32+1+\left(2\right)\left(3\right)

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:\left(2\right)\left(3\right)\::\quad 6

=32+1+6

\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:32+1+6\::\quad 39

=39

Therefore,

4d^3+2e\div \:\:f+de=4\left(2\right)^3+2\left(3\right)\div \:6+\left(2\right)\left(3\right)=39

7 0
4 years ago
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