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Sophie [7]
3 years ago
12

Paul runs 4.8 km every morning. How many feet are in 4.8 km, given that 1 mile = 1.609 km and 1 mile = 5,280 feet?

Mathematics
2 answers:
never [62]3 years ago
7 0
There are <span>15 748.0315 feet  in 4.8 km.
Hope this helps!</span>
attashe74 [19]3 years ago
4 0

Answer:

15.751ft

Step-by-step explanation:

This is a conversion of units problem, so first you have 4.8 km and you need to convert it to feet.

First you take the quantity you need to convert:

4.8km

Then you put a factor, in the denominator you put the conversion factor with the same units that the number that you want to convert, that is:

4.8km*\frac{}{1.609km}

Then in the numerator you put the second unit that you have, in this case, the problem gives you that 1 mile = 1.609km, so:

4.8km*\frac{1mile}{1.609km}

Then, as the problem says that convert the 4.8km to feet, you put the next factor in the same way:

4.8km*\frac{1mile}{1.609km}*\frac{5280feet}{1mile}

Finally, you multiply the initial quantity and the numerators and divide by the denominators, and you have:

4.8km*\frac{1mile}{1.609km}*\frac{5280feet}{1mile}=\frac{25344}{1.609}ft

\frac{25344}{1.609}=15.751ft

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3 years ago
Find a factorization of x² + 2x³ + 7x² - 6x + 44, given that<br> −2+i√√7 and 1 - i√/3 are roots.
Levart [38]

A factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

<h3>What are the properties of roots of a polynomial?</h3>
  • The maximum number of roots of a polynomial of degree n is n.
  • For a polynomial with real coefficients, the roots can be real or complex.
  • The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if a+ib is a root, then a-ib is also a root.

If the roots of the polynomial p(x)=ax^4+bx^3+cx^2+dx+e are r_1,r_2,r_3,r_4, then it can be factorized as p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4).

Here, we are to find a factorization of p(x)=x^4+2x^3+7x^2-6x+44. Also, given that -2+i\sqrt{7} and 1-i\sqrt{3} are roots of the polynomial.

Since p(x)=x^4+2x^3+7x^2-6x+44 is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, -2-i\sqrt{7} and 1+i\sqrt{3} are also roots of the given polynomial.

Thus, all the four roots of the polynomial p(x)=x^4+2x^3+7x^2-6x+44, are: r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}.

So, the polynomial p(x)=x^4+2x^3+7x^2-6x+44 can be factorized as follows:

\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)

Therefore, a factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

To know more about factorization, refer: brainly.com/question/25829061

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Step-by-step explanation:

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