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NeX [460]
3 years ago
15

If the absolute value of x is 4, what is the value of x?

Mathematics
1 answer:
natima [27]3 years ago
4 0
The absolute value of x is 4
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The weight of a boy was 15kg last year. If his weight increases by 3kg this year, find the percentage increase in his weight.
WARRIOR [948]
This year = 18kg
18 x .3= 5.4
The percentage change would be 5.4%. If it needs to be rounded by the nearest tenth, then it would be 5%.
I’m pretty confident this is the answer but I’m not 100% sure. I hope this helps!
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WILL GIVE BRAINLYIST DUE IN 2 HOURS
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Answer:

Step-by-step explanation:

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A group of friends were working on a student film that had a budget of $700. They
Ipatiy [6.2K]

Answer:

231

Step-by-step explanation:

subtract 700 by 469

4 0
3 years ago
Given a function <img src="https://tex.z-dn.net/?f=f%28x%29%3D3x%5E4-5x%5E2%2B2x-3" id="TexFormula1" title="f(x)=3x^4-5x^2+2x-3"
Levart [38]

Answer:

\huge\boxed{f(-1) = -7}

Step-by-step explanation:

In order to solve for this function, we need to substitute in our value of x inside to find f(x). Since we are trying to evalue f(-1), we will substitute -1 in as x to our equation.

f(-1) = 3(-1)^4 - 5(-1)^2 + 2(-1) - 3

Now we can solve for the function by multiplying/subtracting/adding our known values.

Starting with the first term to the last term:

  • 3(-1)^4 = 3

<u><em>WAIT</em></u><em>!</em><em> How is this possible? </em>-1^4 = -1 (according to my calculator), and 3 \cdot -1 = -3, not 3!

It's important to note that taking a power of a negative number and multiplying a negative number are two different things. Let's use -2^2 as an example.

What your calculator did was follow BEMDAS since it wasn't explicitly told not to.

BEMDAS:

- Brackets

- Exponents

- Multiplication/Division

- Addition/Subtraction

Examining the equation, your calculator used this rule properly. Note that exponents come over multiplication.

So rather than  being <em>"-2 squared"</em> - it's <em>"the negative of of 2 squared."</em>

Tying this back into our problem, the squared method would only be true if it looks like -1^4. However, since we're substituting in -1, it looks like (-1)^4, so the expression reads out as "<u><em>-1 to the fourth.</em></u>"

MULTIPLYING -1 by itself 4 times results in -1\cdot-1\cdot-1\cdot-1=1.

Applying this logic to our original term, 3(-1)^4:

  • 3(-1\cdot-1\cdot-1\cdot-1)
  • 3(1)
  • 3

Therefore, our first term is 3.

Let's move on to our second and third terms.

Second term: -5x^2

  • -5(-1)^2

Applying the same logic from our first term:

  • -5(-1 \cdot -1)
  • -5(1)
  • -5

Third term: 2x

  • 2(-1) = -2

-3 is just -3, no influence of x.

Combining our terms, we have 3-5-2-3.

This comes out to be -7, hence, the value of f(-1) for our function f(x)=3x^4-5x^2+2x-3 is <u>-7</u>.

Hope this helped!

4 0
3 years ago
Read 2 more answers
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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