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dedylja [7]
3 years ago
9

0.87 written as a fraction is _____. 8/9 81/90 79/90

Mathematics
2 answers:
Kamila [148]3 years ago
6 0
87/100 Because it’s 87/100
Irina-Kira [14]3 years ago
6 0

Answer:

<em>If the number is 0.87: </em>None of the given options are true.

<em>If the number is 0.87 with the 7 repeating</em>: 79/90 will be your answer.

8.777777... = 79/90 = 8.777777...

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There are 43 pine trees currently in the park. Park workers will plant more pine trees
Luden [163]

Answer:

20 pine trees

Step-by-step explanation:

No of pine trees today = 63

No of pine trees there = 43

So no of pine trees planted today= 63-43= 20

8 0
3 years ago
The Muller family are on holiday in New Zealand. a. They change some euros (€) and receive $1962 (New Zealand dollars). The exch
Readme [11.4K]

Answer:

a. €1200;$1667.70

Step-by-step explanation:

a. Number of euros

\text{euros} = \$1962 \times \dfrac{\text{1 euro}}{\text{\$1.635}}  = \textbf{1200 euros}

b. Dollars remaining

Dollars on hand                        = $1962.00

Less 15 % spent = 0.15 × 1962 = <u>   -294.30</u>

Balance remaining                   =  $1667.70

6 0
3 years ago
Write each of the mixed numbers below as a fraction greater than one, and write each of the fractions greater
Softa [21]

Answer:

a. 13/3

b. 3 3/4

d. 1 7/8

c. 7/2

Step-by-step explanation:

a. 4 1/3 given

4*3=12 multiply the whole numbers with the denominator

12+1 =13 add the numerator to the product

13/3 put sum over denominator

b. 15/4 given

15/4=3 with a remainder of 3 divide numerator by denominator

3 3/4 rewrite as mixed number

d. 15/8 given

15/8=1 with a remainder of 7 divide numerator by denominator

1 7/8 rewrite as mixed number

c. 3 1/2 given

3*2=6 multiply whole number by denominator

6+1=7 add numerator to product

7/2 put sum over denominator

7 0
3 years ago
The diagram shows the ratio of trees Maeve plants to the trees she harvests.<br> Thank u
Sveta_85 [38]

Answer:

Wednesday is 45 to 18.

Thursday is 55 to 22

Step-by-step explanation:

Wednesday: The original ratio was 5 to 2 And the info tells us that the new ratio is ? to 18. To find the first part we have to find how we got from 2 to 18. 2 * 9 = 18. So to find out the first part you do 5 * 9 = 45. We multiply 5 because Thats the original 1st part. We multiply 9 because that's how we got to 18 in the 2nd part. So the answer is 45 for Wednesday.

Thursday: The original ratio was 5 to 2 And the info tells us that the new ratio is 55 to ?. To find the first part we have to find how we got from 5 to 55. 5 * 11 = 55. So to find out the 2nd part you do 2 * 11 = 55. We multiply 2 because That's the original 2nd part. We multiply 11 because that's how we got to 55 in the 1st part. So the answer is 22 for Thursday.

4 0
3 years ago
Please help fast......
Tom [10]

3)

\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\ 1.AY=BX&\text{1. Given}\\ 2.AB \cong AB&\text{2. Reflexive Property}\\ 3. AD || BC&\text{3. Property of a square}\\ 4. \angle ABE \cong \angle AXB&\text{4. Alternate Interior Angles}\\ 5. \angle BAY \cong \angle BYA&\text{5. Alternate Interior Angles}\\6. \triangle BAX \cong \triangle ABY&\text{6. Angle-Side-Angle Theorem}\\ 7. AX \cong BY&\text{7. CPCTC}\\\end{array}

*************************************************************************************

6)

\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\1. AB=CF&\text{1. Given}\\2.AB+BF=A'F&\text{2. Segment Addition Postulate}\\3.CF+BF=A'F&\text{3. Substitution Property}\\4.CF+BF+BC&\text{4. Segment Addition Postulate}\\5.A'F=BC&\text{5. Transitive Property}\\6. \angle AFE = \angle DBC&\text{6. Given}\\7. EF = BD&\text{7. Given}\\8. \triangle AFE \cong \triangle CBD&\text{8. Side-Angle-Side Theorem}\\\end{array}

*************************************************************************************

7)

\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\\text{1.AC bisects }\angle BAD&\text{1. Given}\\2. \angle BAC \cong \angle DAC&\text{2. Property of angle bisector}\\3.AC = AC&\text{3. Reflexive Property}&4. \angle ACB \cong \angle ACD&\text{4. Property of angle bisector}\\5. \triangle ABC \cong \triangle ADC&\text{5. Angle-Side-Angle Theorem}\\6.BC=CD&\text{6. CPCTC}\\\end{array}

5 0
4 years ago
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