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FromTheMoon [43]
3 years ago
11

Poor health is caused by not exercising

Mathematics
1 answer:
fredd [130]3 years ago
4 0
Also can be caused by the foods you eat aswell
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Vickie made a recipe for 144 fluid ounces of scented candle wax. How many 1-cup candle molds can she fill with the recipe
Gre4nikov [31]

there are 0.125 cups inside 1 fluid ounces

144 / 0.125 = 1152 (divide because you are trying to figure out how many of the volumes of the mold can fit inside the 144 fluid ounces

You can fill 1152 molds with 144 fluid ounces

3 0
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

#SPJ9

3 0
1 year ago
Read 2 more answers
If α and β are the zeroes of the polynomial 2x2 + 3x – 7, then find a polynomial whose zeroes are and .
Drupady [299]

Answer:

\frac{-\sqrt{65} - 3 }{4}, \frac{\sqrt{65} + 3}{4}

Step-by-step explanation:

Since we cannot factor the expression, we must use quadratic formula: x = \frac{-b +/-\sqrt{b^2 - 4ac} }{2a}

Plug in <em>a</em> for 2, <em>b</em> for 3, and <em>c</em> for -7 and you should find your roots.

5 0
3 years ago
Which is an equation of a circle with center (2, -1) that passes through the point (3, 4)?
iren2701 [21]
Equation of circle is (x - a)^2 + (y - b)^2 = r^2; where (a, b) is the center and r is the radius.
r^2 = (2 - 3)^2 + (-1 - 4)^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26
Equation is (x - 2)^2 + (y + 1)^2 = 26
8 0
3 years ago
Help me Please..
Fittoniya [83]
Just general definitons:

a TRInomial has 3 terms (tri means three)
a BInomial has 2 terms (bi means two)
a MONOmial has 1 term (mono means one)

the degree is the highest exponent found in the algebraic expression

so they should be pretty easy to solve with that information, but just in case:

1. trinomial, degree of 4
2. binomial, degree of 3
3. monomial, degree of 2

for the final question, all you have to do is plug in 2 for x, so

(2)^2 - 2(2) + 1

4 - 4 + 1

so the answer is 1
7 0
3 years ago
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