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ycow [4]
3 years ago
6

Please help, I didn't learn this in class!

Mathematics
1 answer:
NemiM [27]3 years ago
7 0
Solve the following system using elimination:
{-2 x + 2 y + 3 z = 0 | (equation 1)
{-2 x - y + z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x - 3 y - 2 z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y+0 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)


Divide equation 2 by 5:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:
{-(2 x) + 0 y+3 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
v0 x+0 y+z = 0 | (equation 3)


Subtract 3 × (equation 3) from equation 1:
{-(2 x)+0 y+0 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:
{x+0 y+0 z = 1 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1, y = 1, z = 0

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Help me asap please!!!
Artemon [7]
Well, just by looking at the beginning of the problem, Jamelia had made the common mistake of thinking that 
\sqrt{72} (<span>8.4852...)
</span>is equal to 
2* \sqrt{36} (12)

If you want to estimate a square root like 72, simply find squares that would fit around the number you are looking to find, in our case, 72.

So 9*9 is 81, which is too high and 8*8 is 64, which is too low. So you know that somewhere between those numbers is what your root of 72 is!
5 0
3 years ago
Suppose r(140°, P)(A) = B and (RPD←→∘RPC←→)(A) = B, what is m∠CPD?
NISA [10]

An angle bisector divides an angle into two equal halves.

The measure of angle \angle CPD is 70 degrees

The complete question is an illustrates the concept of angle bisector;

Where:

\angle RPD = 140^o, and line PC bisects \angle RPD

Because line PC bisects \angle RPD, then it means that the measure of RPD is twice the measure of CPD:

So, we have:

\angle RDP = 2 \times \angle CPD

Substitute \angle RPD = 140^o

140^o = 2 \times \angle CPD

Divide both sides by 2

70^o = \angle CPD

Apply symmetric property of equality:

\angle CPD = 70^o

Hence, the measure of angle \angle CPD is 70 degrees

Read more about angle bisectors at:

brainly.com/question/12896755

5 0
2 years ago
Simplify. your answer can only contain positive exponets
Dahasolnce [82]
\frac{4x}{3y}

8 0
3 years ago
Factor the four-term polynomial.
Travka [436]
Group like terms , factor out common variable.

(xz+x) + (yz+y)
x(z+1) + y(z+1)

factor out (z+1)

= (z+1)(x+y)
= solution C.
6 0
3 years ago
Show Your Work! <br> Will Mark Brainliest!
PSYCHO15rus [73]

Answer:

The answer is A.

Step-by-step explanation:

Hope I helped!

6 0
2 years ago
Read 2 more answers
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