Hayami was riding her bike on a loop that is 15/16 mile long. she rode for 3/4 mile and then stopped to rest she rode another 1/8 mile and then has a flat tire. how far does Hayami have to walk her bike to complete the loop?

A.1/16 mile
b.1/32
c. 3/16
d. 1/8

Answers

Answer 1

1/16 mile

3/4 = 12/16

1/8 = 2/16

12/16 + 2/16 = 14/16

1/16 Mile left to walk

Answer 2

Hayami has to walk her bike Option A) 1/16 mile to complete the loop.

Hayami rode a total of 3/4 + 1/8 = 7/8 mile before getting a flat tire. To complete the loop, she needs to cover the remaining distance of the loop, which is

= 15/16 - 7/8

= 15/16 - 14/16

= 1/16 mile.

Thus, Hayami has to walk her bike 1/16 mile to complete the loop. Therefore, the correct answer is 1/16 mile.

To find out how far Hayami has to walk her bike to complete the loop, we need to calculate the remaining distance after she rode her bike before stopping for a rest and getting a flat tire.

1. First, we calculate the distance Hayami rode before stopping for a rest:

  She rode 3/4 mile.

2. Then, we calculate the distance she rode after stopping for a rest but before getting a flat tire:

  She rode 1/8 mile.

3. Now, let's find out the total distance she rode:

  = 3/4 + 1/8

  = 6/8 + 1/8

  = 7/8 mile.

4. The length of the loop is \(15/16\) mile.

5. To find the remaining distance to complete the loop, subtract the distance she rode from the total loop length:

  = 15/16 - 7/8

  = 15/16 - 14/16

  = 1/16 mile.

6. Therefore, Hayami has to walk her bike 1/16 mile to complete the loop.

In conclusion, the correct answer is 1/16 mile (Option A).

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Complete Question

Choose the correct answer from the given options:

Hayami was riding her bike on a loop that is 15/16 mile long. she rode for 3/4 mile and then stopped to rest she rode another 1/8 mile and then has a flat tire. how far does Hayami have to walk her bike to complete the loop?

A.1/16 mile

b.1/32 mile

c. 3/16 mile

d. 1/8 mile


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Answer:

STEP1:

4 Simplify — 1

Equation at the end of step1:

4 (((((((2•(a3))+(4•(a2)))-2a)-————)-(4•(a2)))-4a)+16)•(((((a2)-6a)+(4•a2))+2a)-4) (a3)

STEP2:Equation at the end of step 2

4 (((((((2•(a3))+(4•(a2)))-2a)-————)-(4•(a2)))-4a)+16)•(((((a2)-6a)+22a2)+2a)-4) (a3)

STEP 3 :

Equation at the end of step3:

4 (((((((2•(a3))+(4•(a2)))-2a)-————)-22a2)-4a)+16)•(5a2-4a-4) (a3)

STEP 4 :

4 Simplify —— a3

Equation at the end of step4:

4 (((((((2•(a3))+(4•(a2)))-2a)-——)-22a2)-4a)+16)•(5a2-4a-4) a3

STEP 5 :

Equation at the end of step5:

4 (((((((2•(a3))+22a2)-2a)-——)-22a2)-4a)+16)•(5a2-4a-4) a3

STEP 6 :

Equation at the end of step6:

4 ((((((2a3+22a2)-2a)-——)-22a2)-4a)+16)•(5a2-4a-4) a3

STEP7:Rewriting the whole as an Equivalent Fraction

 7.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  a3  as the denominator :

2a3 + 4a2 - 2a (2a3 + 4a2 - 2a) • a3 2a3 + 4a2 - 2a = —————————————— = ————————————————————— 1 a3

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

STEP8:Pulling out like terms

 8.1     Pull out like factors :

   2a3 + 4a2 - 2a  =   2a • (a2 + 2a - 1) 

Trying to factor by splitting the middle term

 8.2     Factoring  a2 + 2a - 1 

The first term is,  a2  its coefficient is  1 .

The middle term is,  +2a  its coefficient is  2 .

The last term, "the constant", is  -1 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -1 = -1 

Step-2 : Find two factors of  -1  whose sum equals the coefficient of the middle term, which is   2 .

     -1   +   1   =   0

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Adding fractions that have a common denominator :

 8.3       Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

2a • (a2+2a-1) • a3 - (4) 2a6 + 4a5 - 2a4 - 4 ————————————————————————— = ——————————————————— a3 a3

Equation at the end of step8:

(2a6+4a5-2a4-4) (((———————————————-22a2)-4a)+16)•(5a2-4a-4) a3

STEP9:

Rewriting the whole as an Equivalent Fraction :

 9.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  a3  as the denominator :

22a2 22a2 • a3 22a2 = ———— = ————————— 1 a3

STEP10:

Pulling out like terms :

 10.1     Pull out like factors :

   2a6 + 4a5 - 2a4 - 4  = 

  2 • (a6 + 2a5 - a4 - 2) 

Checking for a perfect cube :

 10.2    a6 + 2a5 - a4 - 2  is not a perfect cube

Trying to factor by pulling out :

 10.3      Factoring:  a6 + 2a5 - a4 - 2 

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  -a4 - 2 

Group 2:  a6 + 2a5 

Pull out from each group separately :

Group 1:   (a4 + 2) • (-1)

Group 2:   (a + 2) • (a5)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

 10.4    Find roots (zeroes) of :       F(a) = a6 + 2a5 - a4 - 2

Polynomial Roots Calculator is a set of methods aimed at finding values of  a  for which   F(a)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  a  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  -2.

 The factor(s) are:

of the Leading Coefficient :  1

 of the Trailing Constant :

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