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Anuta_ua [19.1K]
3 years ago
10

Write 5 1/3% as a fraction in simplest form

Mathematics
2 answers:
goblinko [34]3 years ago
7 0
5+1/3=                                                                                                              15/3+1/3=                                                                                                        16/3
choli [55]3 years ago
7 0
In order to find 5 1/3% as a fraction, first convert to an improper fraction...

5 1/3% = 16/3%

To convert % to a fraction we divide by 100...

16/3% = 16/3 ÷ 100

To divide fractions we multiply the reciprocal of the divisor...

16/3 ÷ 100 =
16/3 × 1/100 =

Now multiply

16/3 × 1/100 = 16/300

Now reduce

16/300 = 8/150 = 4/75

5 1/3% = 4/75
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blagie [28]

<u>Question 4</u>

1) \overline{BD} bisects \angle ABC, \overline{EF} \perp \overline{AB}, and \overline{EG} \perp \overline{BC} (given)

2) \angle FBE \cong \angle GBE (an angle bisector splits an angle into two congruent parts)

3) \angle BFE and \angle BGE are right angles (perpendicular lines form right angles)

4) \triangle BFE and \triangle BGE are right triangles (a triangle with a right angle is a right triangle)

5) \overline{BE} \cong \overline{BE} (reflexive property)

6) \triangle BFE \cong \triangle BGE (HA)

<u>Question 5</u>

1) \angle AXO and \angle BYO are right angles, \angle A \cong \angle B, O is the midpoint of \overline{AB} (given)

2) \triangle AXO and \triangle BYO are right triangles (a triangle with a right angle is a right triangle)

3) \overline{AO} \cong \overline{OB} (a midpoint splits a segment into two congruent parts)

4) \triangle AXO \cong \triangle BYO (HA)

5) \overline{OX} \cong  \overline{OY} (CPCTC)

<u>Question 6</u>

1) \angle B and \angle D are right angles, \overline{AC} bisects \angle BAD (given)

2) \overline{AC} \cong \overline{AC} (reflexive property)

3) \angle BAC \cong \angle CAD (an angle bisector splits an angle into two congruent parts)

4) \triangle BAC and \triangle CAD are right triangles (a triangle with a right angle is a right triangle)

5) \triangle BAC \cong \triangle DCA (HA)

6) \angle BCA \cong \angle DCA (CPCTC)

7) \overline{CA} bisects \angle ACD (if a segment splits an angle into two congruent parts, it is an angle bisector)

<u>Question 7</u>

1) \angle B and \angle C are right angles, \angle 4 \cong \angle 1 (given)

2) \triangle BAD and \triangle CAD are right triangles (definition of a right triangle)

3) \angle 1 \cong \angle 3 (vertical angles are congruent)

4) \angle 4 \cong \angle 3 (transitive property of congruence)

5) \overline{AD} \cong \overline{AD} (reflexive property)

6) \therefore \triangle BAD \cong \triangle CAD (HA theorem)

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8) \therefore \vec{DA} bisects \angle BDC (definition of bisector of an angle)

8 0
2 years ago
The diagram show a 5cm x 6cm x 7cm cuboid pls help you
kotegsom [21]

Answer:

d ≈ 10.5

Step-by-step explanation:

The formula for the diagonal of a retangular prism (cuboid) is

d=\sqrt{l^2+w^2+h^2}

1.) Plug in all the values for l, w, and h

d=\sqrt{5^2+6^2+7^2}

2.) simplify by squaring the terms inside the square root

5 * 5 = 25

6 * 6 = 36

7 * 7 = 49

d=\sqrt{25+36+49}

3.) Add the terms

d=\sqrt{110}

4.) Simplify

d = 10.488

5.) Round to one decimal place

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3 years ago
Jeremy is buying a new bicycle. For the style he likes best, the bicycle shop sells 26-inch and 29-inch frames in blue, red, and
Leno4ka [110]

Answer:

1 out of 6

Step-by-step explanation:

there are 6 different combinations

26in - blue

26in- red

26in- black

29in - blue

29in - red

29in - black

The probability then come out to 1 out of 6

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The set of two-digit primes is {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

Of that list, the following primes are mirror images of each other
13 and 31
17 and 71
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Note: we ignore 11 since 11 flips to 11 which is not distinct from its original

If you're looking for the largest prime of this form, then its 97
If you're looking for the largest gap, then subtract each pair
31-13 = 18
71-17 = 54
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Step-by-step explanation:

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