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vfiekz [6]
3 years ago
11

54 is 120% of what number? Explain please!

Mathematics
2 answers:
Nataly [62]3 years ago
3 0
\displaystyle \text{54 is 120\% of what number} \\ \text{is will be = } \\ \text{of will be  } \cdot \\ \text{what number will be z} \\ ---------- \\ 54=120\% \cdot z \\ \\ 54= \frac{120}{100}\cdot z \\ \\\\ z \cdot  \frac{120}{100}=54  \\ \\ \\ z=54 :  \frac{120}{100} \\ \\ \\ z=54 \cdot \frac{100}{120}  \\ \\ \\ z= \frac{5400}{120} =45
boyakko [2]3 years ago
3 0
54 is 120% of the number 64.8 .

If you put 120% in decimal form it's 1.2 if you do 1.2 and multiply it by 54 you get 64.8.

And to check it if you take 64.8 and divide that by 54 you get 1.2 and which is again equivalent to 120%.

I hope this helps a bit.
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Answer:

4/25

Step-by-step explanation:

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D they all have the same slope. 

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Hi I really need help
denis-greek [22]

To solve this, you can first solve each expression:

35 / 6 = 5.83333

5 + 3/10 = 5.3

5 = 5

35 / 10 = 3.5

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3 = 3

10 * 1/2 = 5

So, the tiles of 5, 15*1/5, and 10*1/2 all equal 5.

The tiles of 35/6 and 5+5/6 equal 5.83333.

The remaining tiles to not have matches.

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Which second-degree polynomial function has a leading coefficient of –1 and root 5 with multiplicity 2?
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A second degree polynomial function has the general form:

                  \displaystyle{f(x)=ax^2+bx+c, where a\neq0.

The leading coefficient is a, so we have a=-1.

5 is a double root means that :

i) f(5)=0,
ii) the discriminant D is 0, where D=b^2-4ac.

Substituting x=5, we have

                                        f(5)=a(5)^2+b(5)+c,

and since f(5)=0, and a is -1 we have:

                                          0=-25+5b+c
thus c=25-5b.


By ii) \displaystyle{b^2-4ac=0.

Substituting a with -1 and c with 25-5b we have:
                                     
          \displaystyle{b^2-4ac=0
          \displaystyle{b^2-4(-1)(25-5b)=0
          \displaystyle{b^2+4(25-5b)=0
          \displaystyle{b^2-20b+100=0
          \displaystyle{(b-10)^2=0
          \displaystyle{b=10 


Finally we find c: c=25-5b=25-50=-25

Thus the function is        \displaystyle{f(x)=-x^2+10x-25


Remark: It is also possible to solve the problem by considering the form

f(x)=-1(x-5)^2 directly.

In general, if a quadratic function has leading coefficient a, and has a root r of multiplicity 2, then its form is f(x)=a(x-r)^2
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