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Leviafan [203]
4 years ago
8

Solve for x. −23(3x−4)+3x=56

Mathematics
2 answers:
Neporo4naja [7]4 years ago
7 0
Open bracket
-69x+92+3x=56
Collect like term
-66x=-36
Divide both side by -66
x=6\11
solniwko [45]4 years ago
7 0

Answer: 6/11

Step-by-step explanation:

Open bracket

-69x+92+3x=56

Collect like term

-66x=-36

Divide both side by -66

x=6\11

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Solve each problem<br> -233 minus (-233)<br> 31 minus (-8)<br> -103 minus (-575)
aliya0001 [1]
  • -233 minus (-233)
  • -233 -(-233)
  • -233 + 233
  • 0

  • 31 minus (-8)
  • 31 - (-8)
  • 31 + 8
  • 39

  • -103 - (-575)
  • -103 + 575
  • 472
4 0
2 years ago
Your payment is $375 per month for 36 months and you made a $2,500 a down payment
eimsori [14]

Answer:

$16,000?

Step-by-step explanation:

375 times 36 is 13500

Add 2500

and you have 1600

thats my guess

5 0
3 years ago
Read 2 more answers
Can somebody solve this?
erastova [34]

Answer:

∠ABC = 46°

Step-by-step explanation:

DB and BC are of same length (radius), making it an isosceles triangle.

The base of the triangle is the same at 23°, a total of 46°.

180° - 46° = 134°

But we need ∠ABC, and since points DBA lie on the circumference, it's a straight line.

Therefore, 180° - 134° = 46°

7 0
3 years ago
As a salesperson, you are paid $50 per week plus $2 per sale. This week you want your pay to be at least $100. What is the minim
Georgia [21]
50+2x=100
-50       -50(subtract 50 from both sides)
       2x=50 (divide both sides by 2)
        x=25
8 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
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