Which equivalent form of g(x)=2x^2-12x-54 is the best form to find the zeros of the function?

#16 and #19 only Please answer both as soon as possible.

Vlad1618 [11]

Answer:

1. it is not a direct variation because the value k is not kept throughout the ordered pairs

2. -a^5 - 2a³ + 2a², leading coefficient is -1

Step-by-step explanation:

1. to find the direct variation, we will use y/x = k with k representing our constant throughout the chart. if every value is equal to k, it is a direct variation

for this problem we will assume that the cup sizes are x and the cost is y

so the ordered pairs are (8, 0.90), (12, 1.35), (16, 1.80), and (24, 2.25)

.90/8 = 0.1125 <-- this is our k value for which every result has to be equal to

1.35/12 = 0.1125

1.80/12 = 0.1125

2.25/24 = 0.09375

since the value k was not kept throughout the ordered pairs, this is not a direct variation

2. the standard form of a polynomial is the exponents in descending order.

we have the following polynomial: 2a² - a^5 -2a³

to write it in standard form, we will go in descending order of the exponents. the largest exponent being first

-a^5 - 2a³ + 2a²

the leading coefficient of the polynomial is -1, as there is a negative sign in front of the newly formed polynomial

4 0

3 years ago

What is the degree of the polynomial below? 4x^3+3x^2+6x+5 A. 0 B. 3 C. 1 D. 2

alexira [117]

Answer:

B. 3

Step-by-step explanation:

The degree of this polynomial is based on the highest power on the exponents (if there is more than one variable, it is based on the sum)

The highest power is 3, so the degree is 3

6 0

2 years ago

Read 2 more answers

If ln(4x+y)=2x-3,then dy/dx=

Viktor [21]

$\ln(4x+y)=2x-3$

Differentiate both sides wrt $x$ :

$\dfrac{\mathrm d(\ln(4x+y))}{\mathrm dx}=\dfrac{\mathrm d(2x-3)}{\mathrm dx}$

By the chain rule, we get

$\dfrac1{4x+y}\dfrac{\mathrm d(4x+y)}{\mathrm dx}=2$

$\dfrac{4+\frac{\mathrm dy}{\mathrm dx}}{4x+y}=2$

Solve for $\frac{\mathrm dy}{\mathrm dx}$ :

$4+\dfrac{\mathrm dy}{\mathrm dx}=8x+2y$

$\boxed{\dfrac{\mathrm dy}{\mathrm dx}=8x+2y-4}$

4 0

2 years ago

Determine the volume of the parallelepiped with one vertex at the origin and the three vertices adjacent to it at (2, â1, â1), (

valentinak56 [21]

Answer:

23

Step-by-step explanation:

Here is the complete question

Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at (1, 0, -3), (1, 2, 4), and (5, 1, 0).

Solution

We find the volume of the parallelepiped by making a 3 × 3 column matrix whose columns are the corresponding coordinates of the vertices of the parallelepiped.

So, (1, 0, -3), (1, 2, 4) and (5, 1, 0)

$A = \left[\begin{array}{ccc}1&1&5\\0&2&1\\-3&4&0\end{array}\right]$

The determinant of A is the volume of the parallelepiped. So,

detA = 1(2 × 0 - 4 × 1) - 1(0 × 0 - (-3) × 1) + 5(0 × 4 - (-3) × 2)

= 1(0 - 4) - 1(0 + 3) + 5(0 + 6)

= 1(-4) - 1(3) + 5(6)

= -4 - 3 + 30

= 23

So the volume of the parallelepiped is 23

4 0

2 years ago

Are these numbers reciprocals? a 2 and 1 1/2

faltersainse [42]

Answer:

no they are not reciprocal

Step-by-step explanation:

an example is the reciprocal of 2 would be 1/2

5 0

3 years ago