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Levart [38]
2 years ago
11

Pls help me i will mark as brilliantpls write the step​

Mathematics
1 answer:
djyliett [7]2 years ago
4 0

1a) Firstly, lets start by expanding the brackets. x × x is x². So x(x+1) = x² + x. Then we do the second part. 2 × x = 2x and 2 × 1 = 2. Now we have

x² + 3x + 2.

Now what we do is simplify through quadratic formula. We look at the last digit which is 2. What two numbers multiplies to give you 2? 2 and 1. 2 x 1 = 2 and 2 + 1 = 3. Our second number is 3 so this is the right amount. so now we just put the two factors together. (x + 1)(x+2). This is the answer.

b) I have explained above and so am just gonna write the answers. After expanding we get x² + 2x + 1. We get simplify to

= (x + 1)(x+1)

c) y² + 3y + 4. This cannot be simplified into the format so we leave it like this.

d) = 0.

Have a go at the last ones by yourself, if you still need help, I am available!

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A = wl

Simply divide both sides by w. This is done to isolate the variable we are solving for.

A/w = (wl)/w

A/w = l

Answer: Choice C

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Round to the nearest whole number 486.3÷8.1
PolarNik [594]

Answer: 60.


Step-by-step explanation: After you divide you get 60.037037. And since were rounding its 60.


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3 years ago
Determine algebraically whether the function is even, odd, or neither even nor odd. f as a function of x is equal to 14 times th
svlad2 [7]
Replace x with -x
if you get same function, function is even
if get negative of original, function is odd
if niehter, then neither

if replace with -x
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answer is it is odd function
6 0
3 years ago
This is Matrix for pre calc
gavmur [86]

The given system of equations in augmented matrix form is

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]

If you need to solve this, first get the matrix in RREF:

  • Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]

  • Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]

  • Add 164(row 3) to -91(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]

  • Multiply row 4 by 1/13080:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]

  • Add -153(row 4) to row 3:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]

  • Multiply row 3 by -1/91:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add 6(row 3) and -8(row 4) to row 2:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 2 by 1/5:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 1 by 1/3:

\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

So the solution to this system is (w,x,y,z)=(1,-2,4,-3).

6 0
3 years ago
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