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3 years ago

Carol walked 1 1/3 miles on Monday. 35/6 miles on Tuesday, and 2 miles on Wednesday. About how far did she walk altogether?

Mathematics

2 answers:

valentinak56 [21]3 years ago

5 0

The answer is 9 1/6 because...

Dmitry_Shevchenko [17]3 years ago

4 0

Answer:

9 1/6 miles

Step-by-step explanation:

Add the mixed number by the improper fraction.

1 + 1/3 + 35/6

Solve the fractions first.

In order to have both of the fractions have the same denominator, find the Least Common Multiple of both of the fractions.

1/3 = 2/6

2/6 + 35/6 = 37/6

Turn the improper fraction into a mixed number by dividing the the numerator by the denominator. When you get your quotient, use the remainder as the new numerator over the denominator.

37/6 = 6 1/6

Now, add the 1.

6 1/6 + 1 = 7 1/6

Now, add the 2 miles that Carol walked on Wednesday.

7 1/6 + 2 = 9 1/6

So, Carol walked about 9 1/6 miles on Monday, Tuesday, and Wednesday all together.

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Kipish [7]

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3 years ago

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Alenkasestr [34]

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3 years ago

Statements :

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Answer:

" $e^x$ is irrational for every nonzero integer x"

Step-by-step explanation:

The original statement is

" $e^x$ is rational for some nonzero integer x."

The negation is technically:

"It is NOT true that $e^x$ is rational for some nonzero integer x."

So it's expressing that it's false that $e^x$ can be rational for some nonzero integer x.

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2 years ago

write a polynomial function of least degree with integral coefficients that has the given zeros. -(1/3), -i

inessss [21]

Answer:

$f(x)=3x^3+x^2+3x+1$

Step-by-step explanation:

If a real number $-\frac{1}{3}$ is a zero of polynomial function, then

$x-\left(-\dfrac{1}{3}\right)=x+\dfrac{1}{3}$

is the factor of this function.

If a complex number $-i$ is a xero of the polynomial function, then the complex number $i$ is also a zero of this function and

$x-(-i)=x+i\ \text{ and }\ x-i$

are two factors of this function.

So, the function of least degree is

$f(x)=\left(x+\dfrac{1}{3}\right)(x+i)(x-i)=\left(x+\dfrac{1}{3}\right)(x^2-i^2)=\\ \\ =\left(x+\dfrac{1}{3}\right)(x^2+1)=\dfrac{1}{3}(3x+1)(x^2+1)=\dfrac{1}{3}(3x^3+x^2+3x+1)$

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$f(x)=3x^3+x^2+3x+1$

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3 years ago

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frez [133]

When you evaluate the problem you end up with 204100
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8 0

3 years ago