1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
tia_tia [17]
3 years ago
12

Simone takes part in a 12 1/2 mile race to raise money for charity. She stops 1/8 of the way to rest. How much farther must Simo

ne walk to finish the race?
Mathematics
1 answer:
Vika [28.1K]3 years ago
6 0
Answer 10.4 miles
Explanation:
12.5 divided by 8 = 1.6 miles
12.5-1.6=10.4 miles
You might be interested in
Use a table of function values to approximate an x-value in which the exponential function exceeds the polynomial function. In y
bekas [8.4K]
F(x) = 2^x; h(x) = x^3 + x + 8

Table


x      f(x) =  2^x             h(x) = x^3 + x + 8

0      2^0 = 1                0  + 0 + 8 = 8

1     2^1 = 2                 1^3 + 1 + 8 = 10

2      2^2 = 4                2^3 + 2 + 8 = 8 + 2 + 8 = 18

3      2^3 = 8                3^3  + 3 + 8 = 27 + 3 + 8 = 38

4      2^2 = 16              4^3 + 4 + 8 = 76

10    2^10 = 1024       10^3  +10 + 8 = 1018

 9      2^9 =   512        9^3 + 9 + 8 = 729 + 9 + 8 = 746

Answer: an approximate value of 10
7 0
3 years ago
Jared has total of 16 coins in his piggy bank consisting of only quarters and dimes the total value of all those coins is $3.10
4vir4ik [10]
The answer to the question

4 0
3 years ago
For what value of constant c is the function k(x) continuous at x = 0 if k =
nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x)  is continuous, we take the limit of the parameter as follows:

\mathbf{ \lim_{x \to 0^-} k(x) =  \lim_{x \to 0^+} k(x) =  0 }

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= c }

Provided that:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }

Using l'Hospital's rule:

\mathbf{\implies  \lim_{x \to 0} \ \  \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}=  \lim_{x \to 0}   sec \ x  \ tan \ x = 0}

Therefore:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}=0 }

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

5 0
2 years ago
2/15 plus 3/15 simplify
nikdorinn [45]
2/15+3/15=5/15 5  can go into both 5/5=1 and 15/5=3 which means
=1/3 is the answer 
7 0
3 years ago
Read 2 more answers
If f(x) = 2x - 6 and g(x) = 3x + 9, find (f + g)(x)
alex41 [277]

Answer:

5x+3

Step-by-step explanation:

(f + g)(x) = f(x) + g(x) \\  =2x - 6 + 3x + 9 \\  = 5x + 3

consider marking as Brainliest if this helped

5 0
3 years ago
Other questions:
  • Can someone tell me if this is perpendicular? Thanks!!!
    9·2 answers
  • MIDDLE SCHOOL MATH <br>help
    12·2 answers
  • Need answers for 70-75
    9·1 answer
  • 10. DESIGN Mr. Hagarty is choosing tiles to design a kitche backsplash for a rectangular area 65 inches wide and 9 in tall . Des
    8·1 answer
  • For positive acute angles A and B, it is know that tan A= 8/15 and sin B= 11/61. Find the value of cos (a+b) in simplest form
    13·1 answer
  • Choose 3 values that would make thi inequalitie true 28+x&gt;42
    9·1 answer
  • What is the area of this figure, to the nearest unit?
    8·1 answer
  • The latitude on a map that represents x-axis is the_____________<br><br> (not multiple choice)
    6·2 answers
  • Ayden wants to sell a total of $150 worth of candy bars for his school fundraiser. Each candy bar costs $1.50. How many candy ba
    8·1 answer
  • (√2)^3<br><img src="https://tex.z-dn.net/?f=%20%28%5Csqrt%7B2%29%20%7B%3F%7D%5E%7B3%20%5C%5C%7D%20" id="TexFormula1" title=" (\s
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!