Whats ur favourite color?

If the discriminant, b^2-4ac is ___, a quadratic will have two real roots, two points of intersection with the x-axis.

den301095 [7]

If the discriminant, b² - 4ac is<u> positive</u>

<h3>How to complete the statement?</h3>

From the question, we have the following equation of discriminant

The discriminant (d) is calculated as

d = b² - 4ac

The solutions of a quadratic equation are dependent on the following conditions

If d = 0, the quadratic equation has 1 real solution
If d < 0, the quadratic equation has imaginary solutions
If d > 0, the quadratic equation has 2 different real solutions

This means that "d > 0, the quadratic equation has 2 different real solutions" implies that the discriminant is positive

Hence, the complete statement is (b) positive

Read more about discriminant at

brainly.com/question/64103

#SPJ1

6 0

1 year ago

A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and

nasty-shy [4]

Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

so

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

so

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

3 0

3 years ago

Expand the given power by using Pascal’s triangle. (9a - 10b)^6

xxMikexx [17]

Answer:

$531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

Step-by-step explanation:

1 n=0

1 1 n=1

1 2 1 n=2

1 3 3 1 n=3

1 4 6 4 1 n=4

1 5 10 10 5 1 n=5

1 6 15 20 15 6 1 n=6

This is where n is the exponent in

$(x+y)^n$ .

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

Now we want to expand:

$(9a-10b)^6$ or we we can rewrite as $(9a+(-10b))^6$ .

Let's replace $x$ with $(9a)$ and $y$ with $(-10b)$ in the expansion:

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

$((9a)+(-10b))^6$

$=1(9a)^6+6(9a)^5(-10b)+15(9a)^4(-10b)^2+20(9a)^3(-10b)^3+15(9a)^2(-10b)^4+6(9a)(-10b)^5+1(-10b)^6$

Let's simplify a bit:

$=9^6a^6-60(9)^5a^5b+15(-10)^2(9)^4a^4b^2+20(9)^3(-10)^3a^3b^3+15(9)^2(-10)^4a^2b^4+6(9)(-10)^5ab^5+(-10)^6b^6$

$=531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

8 0

3 years ago

1/4x - 3-4 > 4<br> PLEASE SHOW THE WORK

enot [183]

X>44

.25(x)-3-4>4
.25(x)-7>4
.25(x)>11
x>11

3 0

2 years ago

What is the distance between...

ss7ja [257]

what do you mean ¨ what is the distance between¨ what is the rest of the question?

8 0

3 years ago