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Lady_Fox [76]
2 years ago
9

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>

<h3>⇨ 4y - 7 ✓✓✓ </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(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)

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

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

Now recall the half-angle identities,

cos²(<em>θ</em>/2) = (1 + cos(<em>θ</em>)) / 2

sin²(<em>θ</em>/2) = (1 - cos(<em>θ</em>)) / 2

and taking the positive square roots, we have

cos(<em>θ</em>/2) = √[(1 + cos(<em>θ</em>)) / 2]

sin(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / 2]

Then

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

Notice how we don't need sin(<em>θ</em>) ?

Now, recall the Pythagorean identity:

cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1

Dividing both sides by cos²(<em>θ</em>) gives

1 + tan²(<em>θ</em>) = 1/cos²(<em>θ</em>)

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

cos²(<em>θ</em>) = 1/(1 + tan²(<em>θ</em>))

cos(<em>θ</em>) = - 1/√[1 + tan²(<em>θ</em>)]

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

cos(<em>θ</em>) = - 1/√[1 + (-12/5)²] = - 5/13

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

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

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

3 0
2 years ago
Show work please.<br><br> 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
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