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Fantom [35]
3 years ago
10

What’s the answer? I need help I don’t like similar triangles

Mathematics
2 answers:
Sedbober [7]3 years ago
8 0
I think the answer is 1
Explanation:2x+2=x+3
Subtract the 2 and the x from both sides
1x=1
Divide by one then you get 1
elena-14-01-66 [18.8K]3 years ago
5 0
I think it is one hope this helps
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3. What is the cost of making 100 burritos?<br> Write your answer in the space below.
mel-nik [20]

Answer:

to answer this question you need to give us the cost to make one burrito

6 0
3 years ago
Find x in this proportion.<br> 5/2x =25/4 <br> x =<br><br> 2/5<br> 5/2<br> 4/5<br> 8/5
Ymorist [56]

Answer:

2/5

Step-by-step explanation:

make it reciprocal

2x/5 = 4/25

cross multiply the numbers

2x × 25 = 4×5

2x= 20/25

2x= 4/5

x = 2/5

4 0
3 years ago
What is the vertex of the function in #5
ziro4ka [17]
Where is the function
5 0
3 years ago
Please see attachment
Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }  

Step-by-step explanation:

<u>Step(i)</u>:-

Given function

                       f(x) = \frac{-x}{2x^{2} +1}     ...(i)

Differentiating equation (i) with respective to 'x'

                     f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2}  }   ...(ii)

                    f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  }

Equating Zero

                   f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                 \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                2 x^{2}-1 = 0

               2 x^{2} = 1

             x^{2}  = \frac{1}{2}

             x = \frac{-1}{\sqrt{2} }  , x = \frac{1}{\sqrt{2} }

<u><em>Step(ii):</em></u>-

Again Differentiating equation (ii) with respective to 'x'

f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4}  }

put

      x = \frac{1}{\sqrt{2} }

f^{ll} (x) > 0

The absolute minimum value at   x = \frac{1}{\sqrt{2} }

<u><em>Step(iii):</em></u>-

The value of absolute minimum value

                         f(x) = \frac{-x}{2x^{2} +1}

                       f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}

         on calculation we get

The value of absolute minimum value = - 0.3536      

<u><em>Final answer</em></u>:-

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }    

3 0
3 years ago
Given x is less than y compare the following expressions and determine which is greater 2 x - y or 2y - x explain your answer
aleksandrvk [35]
2y-x is greater. x<y therefore 2x < 2y
6 0
3 years ago
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