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zysi [14]
3 years ago
14

Help plz 20 points do all work

Mathematics
2 answers:
inysia [295]3 years ago
4 0

Answer:

1.7x+1y

2.9x+8n

3.-1r-4n+9y

4.17y-9x

5.-9n-r

Step-by-step explanation:

Combine Like Terms!

Hope this helps! :)

elena-14-01-66 [18.8K]3 years ago
4 0

Answer:

1. 8x + 3y - x - 2y

= 8x - x + 3y - 2y

= 7x + y

2. 3n - 7n - 2n + x + 8x

= 3n - 9n + 9x

= - 6n + 9x

3. 2r - 5n + 9y - 3r + n

= 2r - 3r - 5n + n + 9y

= -r - 4n + 9y

4. 12y - 6x + 5y - 3n

= 12y + 5y - 6x - 3n

= 17y - 9n

5. - 13n + 4n - r

= -9n - r

6. 7xy - 8x + 6xy - y

= 7xy + 6xy - 8x - y

= 13xy - 8x - y

7. - 14r - 10n - 3r - 9n

= -14r - 3r - 10n - 9n

= - 17r - 19n

8. 6r + 9r - 2r - 13n

= 6r + 7r - 13n

= 13r - 13n

9. n + r + x - 3x +4r

= n + r + 4r + x - 3x

= n + 5r - 2x

10. 15y - 7x + 9z + 4z - 3x

= 15y - 7x - 3x + 9z + 4z

= 15y - 10x + 13z

Step-by-step explanation:

You just have to group like terms and then 'add' or 'subtract'.

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attashe74 [19]

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Step-by-step explanation:

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6 0
3 years ago
Help please thank you
stich3 [128]

Answer:3

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8 0
4 years ago
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Help please.......... I need answers
Vikentia [17]

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Step-by-step explanation:

  1. <em><u>7</u></em><em><u>8</u></em>
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6 0
3 years ago
The distance for jasons house to school 0.9 kilometers.what is the distance in meters?
vovikov84 [41]
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5 0
3 years ago
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Use the substitution u = tan(x) to evaluate the following. int_0^(pi/6) (text(tan) ^2 x text( sec) ^4 x) text( ) dx
Rudiy27
If we use the substitution u = \tan x, then du = \sec^2 {x}\ dx. If you try substituting just u and du into the integrand, though, you'll notice that there's a \sec^2x left over that we have to deal with.

To get rid of this problem, use the identity \tan^2 x + 1 = \sec^2 x and substitute in the left side of the identity for the extra \sec^2x, as shown:

\int\limits^{\pi/6}_0 {tan^2 x \ sec^4 x} \, dx
\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx

From there, we can substitute in u and du, and then evaluate:

\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx
\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^2(u^2 + 1)} \, du
\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^4 + u^2} \, du
= \left.\frac{u^5}{5} + \frac{u^3}{3}\right|_0^\frac{1}{\sqrt{3}}
= (\frac{(\frac{1}{\sqrt{3}})^5}{5} + \frac{(\frac{1}{\sqrt{3}})^3}{3}) - (\frac{(0)^5}{5} + \frac{(0)^3}{3})
= \frac{1}{45\sqrt{3}} + \frac{1}{9\sqrt{3}} = \frac{6}{45\sqrt{3}} = \bf \frac{2}{15\sqrt{3}}


8 0
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