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myrzilka [38]
3 years ago
6

Anna hikes at a constant speed travels 6 miles in two hours find her speed in miles per hour you seen an equation for the distan

ce D that Anna travels and H hours
Mathematics
1 answer:
harina [27]3 years ago
3 0

Answer:

The speed at which Anna travel with her bike is 3 miles per hour .

Step-by-step explanation:

Given as :

The distance cover by Anna = 6 miles

The time taken by the Anna to cover the distance = 2 hours

Let The speed in miles per hour = s miles per hour

<u>Now, As we know </u>

Speed = \dfrac{\textrm Distance}{\textrm Time}

So, s =  \dfrac{\textrm 6 miles}{\textrm 2 hours}

or, s = 3 miles per hours

So, Speed = 3 mph

Hence The speed at which Anna travel with her bike is 3 miles per hour . Answer

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Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai
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f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1

Lets check with every option

(a) [-4,-3]

We plug in -4  for x  and -3 for x

f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3  for x  and -2 for x

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.  

(c) [-2,-1]

We plug in -2  for x  and -1 for x

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1  for x  and 0 for x

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0  for x  and 1 for x

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.  

(f) [1,2]

We plug in 1  for x  and 2 for x

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1],  (d) [-1,0], (f) [1,2]

8 0
3 years ago
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Answer:

Step-by-step explanation:

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Two numbers have a product of 48 and a sum of 19. What are the numbers??
Elis [28]
16 * 3 = 48

and 

16 + 3 = 19
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3 years ago
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What is the probability of getting either blue or green on a spinner that is 3/10 greens and 1/5 blue?
Semmy [17]

Answer:

1/2 or 0.50

Step-by-step explanation:

To find the probability of getting either blue or green on a spinner, we must add both probabilities together.

\frac{3}{10}+\frac{1}{5}  = ?

To make adding our two fractions easier, we need to make sure they have like denominators.

\frac{1*2}{5*2} =\frac{2}{10}

Now that the denominators of our probabilities are the same, we can add them together to get our answer.

\frac{3}{10} +\frac{2}{10}=\frac{5}{10}

\frac{5}{10} =\frac{1}{2} or 0.50

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3 years ago
a book store contains 75,972 books, divided in three sections: science, business, and math. thr science section represents 1/3 o
Pachacha [2.7K]

Given that total number of books in the book store = 75972

Given that total number of books in science section =\frac{1}{3} rd of total books =\frac{1}{3}*75972 = 25324

Then remaining number of books = 75972 - 25324 = 50648


Given that 75% of the remaining books are in business section = 75% of 50648 = 0.75*50648 = 37986

Then number of books in the math section = 50648 - 37986 =  12662


Hence final answer is 12662 books in the math section.

4 0
3 years ago
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