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klemol [59]
4 years ago
8

Solve the compound inequality 9 + 3n >= 6 or 5n < 25.

Mathematics
1 answer:
babymother [125]4 years ago
4 0
9+3n \geq 6 \ \lor \ 5n\ \textless \ 25 \\&#10;3n \geq 6-9 \ \lor \ n \ \textless \ \frac{25}{5} \\&#10;3n \geq -3 \ \lor \ n \ \textless \ 5 \\&#10;n \geq \frac{-3}{3} \\&#10;n \geq -1 \\ \\&#10;\boxed{n \geq -1 \hbox{ or } n\ \textless \ 5} \\ \\&#10;\hbox{answer C}
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2x – 3 &gt; 2(x-5) <br> What’s the answer
bulgar [2K]

Answer:

N/A

Step-by-step explanation:

2x - 3 > 2(x-5)

2x - 3 > 2x - 10

2x - 2x > -10 + 3

0 > -7

not true, N/A

5 0
3 years ago
A dance studio offers ballet and hip-hop classes. The studio has 10 ballet classes for every 15 hip hop-hop classes.
liberstina [14]

3 hip hip classes for each ballet class.

8 0
3 years ago
Read 2 more answers
What is the equation of the circle with a center of (8,-2) and a point on the circle of (3,-6)
Artyom0805 [142]

Answer:

(x-3)^2+(y+6)^2=41

Step-by-step explanation:

3 0
3 years ago
Consider the surface defined by the equation x3 + y3 + z3 = 3xyz. Use implicit differ- ∂z entiation to find ∂x. (Note z is not a
Aloiza [94]

Answer:    

   \frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}

Step-by-step explanation:

Given equation is x^{3}+y^{3}+z^{3}=3xyz ………………(1)

we use derivative formula \frac{d}{dx}(x^{n})  = n x^{n-1}

\frac{d}{dx}(x^{3)}  = 3 x^{2}

\frac{d}{dx}(z^{3)}  = 3 z^{2}

And also apply 'u v' formula

\frac{d}{dx}(uv})  = u\frac{d}{dx}(v)+v\frac{d}{dx}(u)

Differentiating  equation (1) partially with respective to 'x' , we treated 'y' as constant.

(3x^{2} +0+3z^{2}\frac{∂z}{∂x}) =3y(x\frac{∂z}{∂x}+z(1))   ( here y treated as constant so the derivative of constant function is zero in addition but in multiplication the constant is keep as like 'y').

on simplification , we get

(3x^{2} +0+3z^{2}\frac{∂z}{∂x}) =(3yx\frac{∂z}{∂x}+3yz)

again simplification, we get

3z^{2}\frac{∂z}{∂x}- 3yx\frac{∂z}{∂x}=3yz-3x^{2}

taking common '\frac{∂z}{∂x} on left on side , we get

(3z^{2}- 3yx)\frac{∂z}{∂x}=3yz-3x^{2}

dividing '(3z^{2}- 3yx) on both sides, we get

\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}

<u>Final answer</u>:-

\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}

6 0
4 years ago
PLEASE HELP QUICKLY IF YOU CAN!
Anastasy [175]

Answer:

6 - (\frac{2}{3} )^{2} =  \frac{50}{9}

3² - (\frac{2}{3} )= 8\frac{1}{3}

7 - (\frac{1}{2} )^{3} = \frac{55}{8}

Step-by-step explanation:

1. 6 - (\frac{2}{3} )^{2}

First, apply the exponent to 2/3. Because it's to the second power, multiply 2/3 by itself.

(\frac{2}{3} )^{2}=\frac{4}{9}

6 - \frac{4}{9}

Now, subtract.

6 - \frac{4}{9} = \frac{50}{9}

------------------------------------------

2. 3² - \frac{2}{3}

First, apply the exponent.

3² = 9

9 - \frac{2}{3}

Now, subtract.

9 - \frac{2}{3}=\frac{25}{3}

\frac{25}{3} =8\frac{1}{3}

------------------------------------------

3. 7 - (\frac{1}{2} )^{3}

First, apply the exponent.

(\frac{1}{2} )^{3} = \frac{1}{8}

7 - \frac{1}{8}

Now, subtract.

7 - \frac{1}{8}=\frac{55}{8}

hope this helps!

4 0
2 years ago
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