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Tpy6a [65]
2 years ago
5

Anyone know the answer?

Mathematics
2 answers:
Margarita [4]2 years ago
8 0

Answer: 3/4

Step-by-step explanation:

9/12 divide by 3/3= 4/3

frosja888 [35]2 years ago
4 0

Answer:

3/4

Step-by-step explanation:

3*3=9

3*4=12

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50= p• 0.05•2<br> Solve
Mariana [72]

Answer:

P=500

Step-by-step explanation:

My reasoning is multiply 0.05 and 2 and you get 0.1. Then you divide 0.1 out of 50 and get 500. Too check the answer you take 500 and multiply it by 0.05 and then  multiply it by 2 and you should get an answer of 50 like it says in the equation.

3 0
3 years ago
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Solve the inequality x/5-3&lt;0
Kisachek [45]

Answer:

x < 15

Step-by-step explanation:

\frac{x}{5} - 3 < 0 ( add 3 to both sides )

\frac{x}{5} < 3 ( multiply both sides by 5 to clear the fraction )

x < 15

7 0
3 years ago
For the feasibility region shown below, find the maximum value of the function P=2x+3y.
timofeeve [1]
Corner points in this graph are: ( 0,0 ) ( 0,8 ) ( 5,6 ) and ( 8, 0 ).
If we plug those values in : P = 2 x + 3 y
P ( 0,0 )= 0
P ( 0,8 ) = 2 * 0 + 3 * 8 = 24
P ( 6 , 5 ) = 2 * 6 + 3 * 5 = 12 + 15 = 27
P ( 8 , 0 ) =  2 * 8 + 3 * 0 = 16
The maximum value is:
P max ( 6 , 5 ) = 27
3 0
3 years ago
Read 2 more answers
For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course
pav-90 [236]

By definition of absolute value, you have

f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1

or more simply,

f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For <em>x</em> > -1, we have

(<em>x</em> + 1)<em>'</em> = 1

while for <em>x</em> < -1,

(-<em>x</em> - 1)<em>'</em> = -1

More concisely,

f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x

Note the strict inequalities in the definition of <em>f '(x)</em>.

In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:

\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1

\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1

All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.

4 0
3 years ago
22x(4+16)-78x3
Deffense [45]

Answer:

what we solving for???? ill answer in the comments

Step-by-step explanation:

4 0
2 years ago
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