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horsena [70]
3 years ago
12

1. The top 1/8 of the runners advance to the finals for the track and field competition. What is the decimal value of the number

? Show your work.
Mathematics
2 answers:
Scorpion4ik [409]3 years ago
7 0
1/8=0.125
  
      0.125
    ______
8 I  1.000
     -   8
    ______
         20
     -   16
   _______
           40
     -     40
   _______
             0
alexgriva [62]3 years ago
5 0

Answer:

Hence the number in decimal is:

0.125

Step-by-step explanation:

We are given that:

The top 1/8 of the runners advance to the finals for the track and field competition.

We are asked to find the decimal value of the number i.e. we have to represent 1/8 in the decimal value.

on dividing the number 1 ( which is in numerator) by 8 ( which is in denominator) we obtain:

\dfrac{1}{8}=0.125

Hence the number in decimal is:

0.125

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1. Evaluate: <br>0.234 x 10^2<br><br>​
stellarik [79]

Answer: 23.4 is the answer

3 0
2 years ago
Read 2 more answers
Plz Help
GenaCL600 [577]

Begin by factoring 2 out of   2x^2 - 2x - 12 equals 0:

2(x^2 - x - 6) = 0

2(x - 3)(x + 2) = 0.  2 is never zero, but x-3 and x+2 can each be set = to 0:

This results in x = 3 and x = -2.  The equation is true for these two x-values.

6 0
3 years ago
In a test, John was able to answer 15 correct out of 40 questions. Beth was able to answer 24 questions correctly out of 50. Mic
Helen [10]

Answer:

that is wrong beth has a better score

Step-by-step explanation:

14 out of 40 is 37.5%

and 24 out of 50 is 48%

4 0
2 years ago
Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F
Svet_ta [14]

Answer:

A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36

Step-by-step explanation:

The equations are:

y = x^{2} + 2x + 3

y = 2x + 12

The two graphs intersect when:

x^{2} + 2x + 3 = 2x + 12

x^{2} = 0

x_{1}  = 3\\x_{2}  = -3

To find the area under the curve for the first equation:

A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx

To find the area under the curve for the second equation:

A_{2} = \int\limits^3__-3}{2x + 12} \, dx

To find the total area:

A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx

Simplifying the equation:

A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).  

3 0
3 years ago
How do I do this because it's really confusing haha
netineya [11]
A = 1.5
B = -1/3
C = -4/3

A is a positive so there is only one possibility
B is in between 0 and -1, so it has to be -1/3
C is below -1 and so its the improper fraction

5 0
3 years ago
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