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sp2606 [1]
3 years ago
6

Complete each blank to find a expression that is equal to 16%

Mathematics
2 answers:
Alenkasestr [34]3 years ago
8 0

We will complete the given blanks in the given expressions that is equal to 16%.

A. ____ for every 100 = 16%

Let the unknown value be 'x'

x for every 100 = 16%

\frac{x}{100}=\frac{16}{100}

x = 16

Therefore, 16 for every 100 is 16% that is 0.16

B. ___ for every 50 = 16%

Let the unknown value be 'y'

y for every 50 = 16%

\frac{y}{50}=\frac{16}{100}

\frac{y}{50}=0.16

y = 8

Therefore, 8 for every 50 is 16%.

C. 1 for every ___ is 16%

Let the unknown value be 'z'

1 for every z is 16%

\frac{1}{z}=\frac{16}{100}

100=16z

z=\frac{100}{16}

z=6.25

1 for every 6.25 is 16%.

D. 0.5 for every ___ is 16%

Let the unknown value be 'w'

0.5 for every w is 16%

\frac{0.5}{w}=\frac{16}{100}

50=16w

w=\frac{50}{16}

w = 3.125

1 for every 3.125 is 16%.

beks73 [17]3 years ago
4 0
<span>A. 16 for every 100
 B.  8 for every 50
 C. 1 for every 6.25
 D. 0.5 for every 3.125</span>
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Which method would you use to prove that the two triangles are congruent?
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Answer:- AAS postulate


Explanation:-

  • AAS postulate tells that if two angles and a non-included side of a triangle to equal to the two angles and a non-included side of another triangle then the two triangles are said to be congruent.

Given:- One angle and one side of a triangle is equal to the one angle and one side of the other triangle.

We see there is one more pair of equal angles as they are vertically opposite angles . [See the attachment]

⇒ there is a triangle where two angles and a non-included side of a triangle to equal to the two angles and a non-included side of another triangle then the two triangles are said to be congruent.

⇒ The triangles are congruent [ by ASA postulate]


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3 years ago
A rectangle has vertices at (0 0) (3 0) and (0 6). what is the area of the rectangle answer
KonstantinChe [14]

Answer:

18 sq. units

Step-by-step explanation:

Height would be 6 and Length would be 3

6 x 3 = 18

For further help, use a coordinate grid and plot the points

3 0
3 years ago
Read 2 more answers
PLEASE HELP
Tems11 [23]

QUESTION A

The given multiplication problem is

\frac{39}{64} \times \frac{8}{13}

Factor each term to obtain;

\frac{13\times 3}{8\times8} \times \frac{8}{13}

Cancel out the common factors to obtain;

\frac{1\times 3}{8\times1} \times \frac{1}{1}

Simplify to get;

\frac{3}{8}

QUESTION B

The given multiplication problem is

\frac{2}{3}\times \frac{1}{5}\times \frac{4}{7}

This the same as

\frac{2\times 1\times 4}{3\times 5\times 7}

This simplifies to;

\frac{8}{105}

QUESTION C

The given problem is

\frac{3}{5}\times \frac{10}{12} \times \frac{1}{2}

This is the same as

\frac{3}{5}\times \frac{5}{6} \times \frac{1}{2}

=\frac{1}{1}\times \frac{1}{2} \times \frac{1}{2}

This simplifies to

=\frac{1}{4}

QUESTION D.

The given expression is

\frac{4}{9}\times 54

Factor the 54 to obtain;

\frac{4}{9}\times 9\times 6

Cancel the common factors to get;

\frac{4}{1}\times 1\times 6

This simplifies to;

=24

QUESTION E

The given problem is

20\times 3\frac{1}{5}

Convert the mixed numbers to improper fraction to obtain;

=20\times \frac{16}{5}

=4\times5 \times \frac{16}{5}

Cancel the common factors to get;

=4\times1 \times \frac{16}{1}

=64

QUESTION F

The multiplication problem is

11 \times 2 \frac{7}{11}

Convert the mixed numbers to improper fractions to obtain;

11 \times \frac{29}{11}

Cancel out the common factors to get;

=1 \times \frac{29}{1}

Simplify;

=29

QUESTION G

The given problem is

5\frac{1}{3}\times 5\frac{1}{8}

Convert to improper fractions;

=\frac{16}{3}\times \frac{41}{8}

Cancel out the common factors to get;

=\frac{2}{3}\times \frac{41}{1}

=\frac{82}{3}

Convert back to mixed numbers

=27\frac{1}{3}

QUESTION H

The given expression is

10\frac{2}{3} \times 1\frac{3}{8}

Convert to improper fraction to get;

\frac{32}{3} \times \frac{11}{8}

Cancel common factors to get;

=\frac{4}{3} \times \frac{11}{1}

Simplify

=\frac{44}{3}

Convert back to mixed numbers;

=14\frac{2}{3}

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