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2 years ago

What is the completely factored form of this polynomial? 18x^3– 120x^2-42x

Mathematics

2 answers:

andreyandreev [35.5K]2 years ago

8 0

Hope this helps! feel free to clarify

Vlada [557]2 years ago

4 0

Answer:

6×(3×+1)(×-7)

Step-by-step explanation:

18×^3-120×^2-42×

=6×(3×+1)(×-7)

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PLEASE HELP ME WILL GIVE BRAINLIEST TO CORRECT ANSWER

alex41 [277]

common difference, d = -3

f1 = -13

An arithmetic sequence f(n) = f1 + d(n - 1)

so f(n) = -13 - 3(n - 1)

f(46) = -13 - 3(46-1) = -13 -3(45) = -13 - 135 = -148

Answer:

f(46) = - 148

8 0

2 years ago

Read 2 more answers

Pls help i will give u brianliest

Fynjy0 [20]

Answer:

Slope = -3

y-intercept = (0,-4)

Equation: y = -3x-4

Step-by-step explanation:

We can take any two input-output pairs from the table to find the equation of given function.

Linear function is given by:

$y =mx+b$

Here m is the gradient of the functions which is defined as:

$m = \frac{rise}{run} = \frac{y_2-y_1}{x_2-x_1}$

The input-output pairs are:

(x1,y1) = (-1,-1)

(x2,y2) = (0,-4)

First of all,

$m = \frac{-4-(-1)}{0-(-1)}\\= \frac{-4+1}{1} = -3$

Putting the value of slope in the equation

$y = -3x+b$

Putting (-1,-1) in the equation

$-1 = -3(-1)+b\\-1 = 3+b\\b = -1-3\\b = -4$

The equation will be:

$y = -3x-4$

Hence,

Slope = -3

y-intercept = (0,-4)

Equation: y = -3x-4

7 0

3 years ago

Please help me solve this, I will give you Brainliest.

nadezda [96]

<h3>That means to subtract 7, from 4 times of 'y' .</h3>

6 0

3 years ago

Read 2 more answers

This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, pl

Lostsunrise [7]

First,

tan(θ) = sin(θ) / cos(θ)

and given that 90° < θ < 180°, meaning θ lies in the second quadrant, we know that cos(θ) < 0. (We also then know the sign of sin(θ), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < θ/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(θ/2) > 0 and sin(θ/2) > 0.

Now recall the half-angle identities,

cos²(θ/2) = (1 + cos(θ)) / 2

sin²(θ/2) = (1 - cos(θ)) / 2

and taking the positive square roots, we have

cos(θ/2) = √[(1 + cos(θ)) / 2]

sin(θ/2) = √[(1 - cos(θ)) / 2]

Then

tan(θ/2) = sin(θ/2) / cos(θ/2) = √[(1 - cos(θ)) / (1 + cos(θ))]

Notice how we don't need sin(θ) ?

Now, recall the Pythagorean identity:

cos²(θ) + sin²(θ) = 1

Dividing both sides by cos²(θ) gives

1 + tan²(θ) = 1/cos²(θ)

We know cos(θ) is negative, so solve for cos²(θ) and take the negative square root.

cos²(θ) = 1/(1 + tan²(θ))

cos(θ) = - 1/√[1 + tan²(θ)]

Plug in tan(θ) = - 12/5 and solve for cos(θ) :

cos(θ) = - 1/√[1 + (-12/5)²] = - 5/13

Finally, solve for sin(θ/2) and tan(θ/2) :

sin(θ/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(θ/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2

3 0

2 years ago

Show work please. solve system of equations using matrices.

nadya68 [22]

Answer:

(t, t -1, t)

Step-by-step explanation:

You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called. I think it refers to infinite many solutions). Here's how it works:

Set up your matrix:

$\left[\begin{array}{ccc}1&-2&1\\2&-1&-1\\\end{array}\right] \left[\begin{array}{ccc}2\\1\\\end{array}\right]$

You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left. Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0. Multiplying in a -2 to the top row gives you:

$\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right]$

Then add, keeping the first row the same and changing the second to reflect the addition:

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

The second equation is this now:

3y - 3z = -3. Solving for y gives you y = z - 1. Let's let z = t (some random real number that will make the system true. Any number will work. I'll show you at the end. Just bear with me...)

lf z = t, and if y = z - 1, then y = t - 1. So far we have that y = t - 1 and z = t. Now we solve for x:

From the first equation in the original system,

x - 2y + z = 2. Subbing in t - 1 for y and t for z:

x - 2(t - 1) + t = 2. Simplify to get

x - 2t + 2 + t = 2 and x - t = 0, and x = t. So the solution set is (t, t - 1, t). Picking a random value for t of, let's say 2, sub that in and make sure it works. If:

x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2. Check it: 2 - 4 + 4 = 2 and 2 = 2. You could pick any value for t and it will work.

6 0

2 years ago