Ask question

Ask question

All categories

2 years ago

7

Conjecture help? I need one!

1 answer:

kirill [66]2 years ago

6 0

hii! The answer is false. This is true because if you calculate the average of say, 2 and 4 will be 2+4=6. And we divide that by 2 and we get 3 as an answer. 3 Is NOT an even number. I'm so bored that I'm chilling on brainly now smh. Anyways I hope u have a good night or day! - Castiel (She/They)

You might be interested in

Simplify -8.6 + 2.3 + 1<br><br> -5.3<br> -7.3<br> -9.9

telo118 [61]

Answer:

Hello, The answer is -5.3

Step-by-step explanation:

Have a great day!

8 0

3 years ago

Read 2 more answers

Rectangle ABCD is graphed in the coordinate plane. The following are the vertices of the rectangle:

ArbitrLikvidat [17]

Answer:

5x3

Step-by-step explanation:

7 0

3 years ago

Amy is dressing a mannequin for a store window to show off a

castortr0y [4]

Answer:

8

Step-by-step explanation:

4 different shirts would be 4 combinations. You can have one of each of the jeans on the mannequin, giving 4x2=8.

3 0

3 years ago

Read 2 more answers

Write 90% as a decimal. Round to the thousandths place

dybincka [34]

90% = 90/100 = 9/10 = 0.9
Hope this helps :)

4 0

4 years ago

Read 2 more answers

find the zeros of following quadratic polynomial and verify the relationship between the zeros and the coefficient of the polyno

Cloud [144]

Answer:

$\textsf{Zeros}: \quad x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}$

Step-by-step explanation:

Rewrite the given polynomial in the form ax² + bx + c:

$f(x)=2 \sqrt{3}x^2+5x-4 \sqrt{3}$

To find the zeros, set the function to zero and solve for x using the quadratic formula.

$\implies 2 \sqrt{3}x^2+5x-4 \sqrt{3}=0$

<u>Quadratic formula</u>:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore,

a = 2√3
b = 5
c = - 4√3

Substituting the values into the quadratic formula:

$\implies x=\dfrac{-5 \pm \sqrt{5^2-4(2\sqrt{3})(-4\sqrt{3})} }{2(2\sqrt{3})}$

$\implies x=\dfrac {-5 \pm \sqrt {121}}{4\sqrt{3}}$

$\implies x=\dfrac {-5 \pm 11}{4\sqrt{3}}$

$\implies x=\dfrac {6}{4\sqrt{3}}, \:\:x=\dfrac {-16}{4\sqrt{3}}$

$\implies x=\dfrac {3}{2\sqrt{3}}, \:\:x=-\dfrac {4}{\sqrt{3}}$

$\implies x=\dfrac {\sqrt{3}}{2}, \:\:x=-\dfrac {4\sqrt{3}}{3}$

The sum of the roots of a polynomial is -b/a:

$\implies -\dfrac{b}{a}=-\dfrac{5}{2 \sqrt{3}}=-\dfrac{5\sqrt{3}}{6}$

The sum of the found roots is:

$\implies \left(\dfrac {\sqrt{3}}{2}\right)+\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{5\sqrt{3}}{6}$

Hence proving the sum of the roots is -b/a

The product of the roots of a polynomial is: c/a

$\implies \dfrac{c}{a}=\dfrac{-4\sqrt{3}}{2\sqrt{3}}=-2$

The product of the found roots is:

$\implies \left(\dfrac {\sqrt{3}}{2}\right)\left(-\dfrac {4\sqrt{3}}{3}\right)=-\dfrac{12}{6}=-2$

Hence proving the product of the roots is c/a

Therefore, the relationship between the roots and the coefficients is verified.

8 0

2 years ago