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2 years ago

11

Plese HELP !!!

1 answer:

beks73 [17]2 years ago

6 0

Answer:

-9/8

Step-by-step explanation:

$\frac{-2a^{-3}b^2}{a} \\ \\ =-\frac{2b^2}{a^4} \\ \\ =-\frac{2(3)^2}{(-2)^4} \\ \\ =-\frac{18}{16} \\ \\ =-\frac{9}{8}$

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What is f(5)? According to the graph.

Levart [38]

Answer:

f(5)=-7

Step-by-step explanation:

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3 years ago

What are the roots of 2x^2+10×-48=0

mojhsa [17]

X=-8,3.
divide all terms by the two in the first term to get x^2+5x-24=0. Then find the factors. To do so, find what two numbers can be multiplied to get 24, and whether those two numbers can be added or subtracted to get the middle term (5). Since 8x(-3)=24 and 8-3=5, (x+8) and (x-3) are the factors. Then set the factors equal to zero. Add or subtract these numbers to the other side of the equation and you then get x=-8,3. These are the roots.

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3 years ago

An airline allows each passenger 30

ollegr [7]

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3 years ago

{96/[36/3-(18 x 2 - 30)]} / (31-16+1)

ehidna [41]

Answer:

-1/2

Step-by-step explanation:

pemdas rules

1. 18 x 2 - 30 = 6

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3 0

3 years ago

Determine the number of real solutions for each of the given equations. Equation Number of Solutions y = -3x2 + x + 12 y = 2x2 -

rosijanka [135]

Answer:

Step-by-step explanation:

Our equations are

$y = -3x^2 + x + 12\\y = 2x^2 - 6x + 5\\y = x^2 + 7x - 11\\y = -x^2 - 8x - 16\\$

Let us understand the term Discriminant of a quadratic equation and its properties

Discriminant is denoted by D and its formula is

$D=b^2-4ac\\$

Where

a= the coefficient of the $x^{2}$

b= the coefficient of $x$

c = constant term

Properties of D: If D

i) D=0 , One real root

ii) D>0 , Two real roots

iii) D<0 , no real root

Hence in the given quadratic equations , we will find the values of D Discriminant and evaluate our answer accordingly .

Let us start with

$y = -3x^2 + x + 12\\a=-3 , b =1 , c =12\\D=1^2-4*(-3)*(12)\\D=1+144\\D=145\\D>0 \\$

Hence we have two real roots for this equation.

$y = 2x^2 - 6x + 5\\$

$y = 2x^2 - 6x + 5\\a=2,b=-6,c=5\\D=(-6)^2-4*2*5\\D=36-40\\D=-4\\D$

Hence we do not have any real root for this quadratic

$y = x^2 + 7x - 11\\a=1,b=7,-11\\D=7^2-4*1*(-11)\\D=49+44\\D=93\\$

Hence D>0 and thus we have two real roots for this equation.

$y = -x^2 - 8x - 16\\a=-1,b=-8,c=-16\\D=(-8)^2-4*(-1)*(-16)\\D=64-64\\D=0\\$

Hence we have one real root to this quadratic equation.

7 0

3 years ago