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Ghella [55]
3 years ago
14

Juan delivers newspapers at a daily rate of $30. He also gets an extra $0.25 per newspaper delivered before 8 A.M. If Juan was p

aid $36.25 for Sunday, how many newspapers did he deliver before 8 A.M? *
Mathematics
1 answer:
irina [24]3 years ago
7 0

Answer:

he delivered 25 newspapers be for 8 am

Step-by-step explanation:

have a good day!

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How many solutions are there to the equation below? 4(x-5)=4x-24
blondinia [14]

It is a linear equation so there can be 1 solution, 0 solutions, or infinite solutions

4(x-5)=4x-24

distribute

4x -20 = 4x-24

subtract 4x from each side

-20 = -24

no solutions


6 0
3 years ago
Read 2 more answers
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

#SPJ9

3 0
1 year ago
Read 2 more answers
What is the simplified form of this expression?
Alexxx [7]

Answer:

Option 4

Step-by-step explanation:

=> -3x^2+2x-4 + 4x^2+5x+9

Combining like terms

=> -3x^2+4x^2+2x+5x-4+9

=> x^2+7x+5

5 0
3 years ago
You work 4 hours on Monday, 4.5 hours on Tuesday, 7.25 hours on Thursday, and 12 hours on Saturday. If you get paid $11.52 per h
likoan [24]

Answer:

319.68

Step-by-step explanation:

you multyply 11.52 per each hour and then add everything up

Wish this helped you

7 0
3 years ago
Which expression is equivalent to 3/24?
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5 0
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