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3 years ago

14

maria,leah, and jonas ran these distances on Saturday. maria ran 5/6 miles, leah ran 2/3 mile, jonas ran 3/4 mile. who ran the s

hortest distance?

1 answer:

rjkz [21]3 years ago

7 0

Maria ran 5/6 miles

Leah ran 2/3 mile

Jonas ran 3/4 mile

Order from Least to Greatest
2/3 < 3/4 < 5/6

Leah ran the shortest distance.

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Solve 2(x + 1) = 2x + 5.

BlackZzzverrR [31]

Answer:

No solutions

Step-by-step explanation:

2(x+1)=2x+5

Expand the parentheses:

2x+2=2x+5

Subtract 2x from both sides:

2=5

Since this is not true, this equation has no solutions. Hope this helps!

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jonny [76]

Answer:

If her brother has 40 tickets, the equation will be 3(40)+2=122. jamie will have 122 tickets

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Step-by-step explanation:

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3 years ago

If c (x) = StartFraction 5 Over x minus 2 EndFraction and d(x) = x + 3, what is the domain of (cd)(x)?

Volgvan

Answer:

The domain of the function (cd)(x) will be all real values of x except x = 2.

Step-by-step explanation:

The two functions are $c(x) = \frac{5}{x - 2}$ and d(x) = x + 3

So, (cd)(x) = $(\frac{5}{x - 2})(x + 3) = \frac{5(x + 3)}{x - 2}$

Then, for x = 2 the function (cd)(x) will be undefined as zero in the denominator will make the function (cd)(x) undefined.

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3 years ago

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Paraphin [41]

Answer:

If you want to use the Rational Zeros Theorem, as instructed, you need to use synthetic division to find zeros until you get a quadratic remainder.

P: ±1, ±2, ±3, ±6 (all prime factors of constant term)

Q: ±1, ±7 (all prime factors of the leading coefficient)

P/Q: ±1, ±2, ±3, ±6, ±1/7, ±2/7, ±3/7, ±6/7 (all possible values of P/Q)

Now, start testing your values of P/Q in your polynomial:

f(x)=7x4-9x3-41x2+13x+6

You can tell f(1) and f(-1) are not zeros since they're not = 0Now try f(2) and f(-2):

f(2)=7(16)-9(8)-41(4)+13(2)+6

112-72-164+26+6 ≠ 0

f(-2)=7(16)-9(-8)-41(4)+13(-2)+6

112+72-164-26+6 = 0 OK!! There is a zero at x=-2

This means (x+2) is a factor of the polynomial.

Now, do synthetic division to find the polynomial that results from

(7x4-9x3-41x2+13x+6)÷(x+2):

-2⊥ 7 -9 -41 13 6

-14 46 -10 -6

7 -23 5 3 0 The remainder is 0, as expected

The quotient is a polynomial of degree 3:

7x3-23x2+5x+3

Now, continue testing the P/Q values with this new polynomial. Try f(3):

f(3)=7(27)-23(9)+5(3)+3

189-207+15+3 = 0 OK!! we found another zero at x=3

Now, another synthetic division:

3⊥ 7 -23 5 3

21 -6 -3_

7 -2 -1 0

The quotient is a quadratic polynomial:

7x2-2x-1 This is not factorable, you need to apply the quadratic formula to find the 3rd and 4th zeros:

x= (1±2√2)÷7

The polynomial has 4 zeros at x=-2, 3, (1±2√2)÷7

Step-by-step explanation:

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