4. Michael observed he felt the pain of losing a $20 bill more than he felt the joy of finding it on the sidewalk the

What is the value of the expression below when y=6y=6? y^2 -5y-4 y 2 −5y−4

Elodia [21]

Answer:

2

Step-by-step explanation:

Given expression:

y^2 - 5y - 4

y = 6

Substitute y = 6 into the expression

y^2 - 5y - 4

6^2 - 5(6) - 4

(6*6) - (5*6) - 4

36 - 30 - 4

36 - 34

2

The value of the expression y^2 - 5y - 4 is 2 when y = 6

7 0

3 years ago

the area of a rectangular barn is 96 square feet. its length is 4 feet longer than the width. find the length and width of the b

lana [24]

Answer:

Triangle

Step-by-step explanation:

2=2

5 0

2 years ago

Convert the polar representation of this complex number into its standard form z=4(cos150+isin150).

Katen [24]

Answer:

B

Step-by-step explanation:

Using the exact values

cos150° = - cos30° = - $\frac{\sqrt{3} }{2}$

sin150° = sin30° = $\frac{1}{2}$

Given

z = 4(cos150° + isin150° ) , substitute values

= 4(- $\frac{\sqrt{3} }{2}$ + $\frac{1}{2}$ i )

= - 2 $\sqrt{3}$ + 2i , that is

(- 2 $\sqrt{3}$ , 2 ) → B

7 0

3 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

or

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

or

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

or

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

2 years ago

When simplifying an exponential expression what happens to an exponent when it is moved from the numerator to the denominator or

Gala2k [10]

Answer:

Hi I don't really get it but I'll do my best to figure it out.

Basically if a numerator has a negative exponent it would be moved to the bottom to make the exponent positive.

IF there is a negative exponent at the denominator you move it to the top to make it positive.

Hope that helps.

3 0

3 years ago