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3 years ago

Please help fastit's for a quiz

Mathematics

1 answer:

horsena [70]3 years ago

6 0

Answer:

Step-by-step explanation:

5 pi > 15

$\pi = 3.14$

5 * 3.14 > 15

15.7 > 15

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bekas [8.4K]

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3 years ago

Today 13 athletes were tested for steroids. Yesterday three times as many athletes were tested. How many athletes were tested in

Sauron [17]

13 were tested
today 3 times as much as yesterday was tested
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4 years ago

Help me out guys. Got stuck over this algebra question

vovikov84 [41]

Answer:

Compare LHS and RHS to prove the statement.

Step-by-step explanation:

Given: a + b + c = 0

We have to show that $\frac{1}{1 + x^b + x^{-c}} + \frac{1}{1 + x^c + x^{-a}} + \frac{1}{1 + x^a + x^{-b}} = 1$

We take LCM, simplify the terms and compare LHS and RHS. We will see that LHS = RHS and the statement will be proved.

Taking LCM, we get:

$\frac{(1 + x^c + x^{-a})(1 + x^a + x^{-b}) + (1 + x^b + x^{-c})(x^a + x^{-b} + 1) + (x^b + x^{-c} + 1)(x^c + x^{-a} + 1)}{(1 + x^a + x^{-b})(1 + x^b + x^{-c})(1 + x^c + x^{-a})}$ = 1

⇒ $(1 + x^c + x^{-a})(1 + x^a + x^{-b}) + (1 + x^b + x^{-c})(x^a + x^{-b} + 1) + (x^b + x^{-c} + 1)(x^c + x^{-a} + 1) = (1 + x^a + x^{-b})(1 + x^b + x^{-c})(1 + x^c + x^{-a})$

We simplify each term and then compare LHS and RHS.

Simplifying the first term:

$(1 + x^c + x^{-a})(1 + x^a + x^{-b})$

= $x^{c + a} + x^{c - b} + x^c + 1 + x^{-a - b} + x^{-a} + x^{a} + x^{-b} + 1$

= $x^{c + a} + x^{c - b} + x^{c} + x^{-a - b} + x^{-a} + x^{a} + x^{-b} + 2 \hspace{5mm} \hdots (A)$

Now, we simplify the second term we have:

$(1 + x^b + x^{-c})(x^a + x^{-b} + 1)$

= $x^{a + b} + 1 + x^{b} + x^{a - c} + x^{-b - c} + x^{-c} + x^{a} + x^{-b} + 1$

= $x^{a + b} + x^{b} + x^{a - c} + x^{- c - b} + x^{-c} + x^a + x^{-b} + 2 \hspace{5mm} \hdots (B)$

Again, simplifying $(x^b + x^{-c} + 1)(x^c + x^{-a} + 1)$ ,

= $x^{b + c} + x^{b - a} + x^b + 1 + x^{-c -a} + x^{-c} + x^c + x^{-a} + 1$

= $x^{b + c} + x^{b - a} + x^b + x^{-c -a} + x^{-c} + x^c + x^{-a} + 2$ (C)

Therefore, LHS = A + B + C

= $x^{c + a} + x^{c - b} + 2x^c + x^{-a - b} + 2x^{-a} + 2x^{a} + 2x^{-b} + x^{a + b} + 2x^b + x^{a - c} + x^{-c -b} + x^{b + c} + x^{b - a} + x^{-c -a} + 6$

Similarly. RHS

= $(x^{b + c} + x^{b - a} + x^b + 1 + x^{-c -a} + x^{-c} + x^c + x^{-a} + 1)(x^a + x^{-b} + 1)$

Note that if a + b + c = 0, $\implies x^{a + b + c} = x^0 = 1$

So, RHS = $x^{c + a} + x^{c - b} + 2x^c + x^{-a - b} + 2x^{-a} + 2x^{a} + 2x^{-b} + x^{a + b} + 2x^b + x^{a - c} + x^{-c -b} + x^{b + c} + x^{b - a} + x^{-c -a} + 6$

We see that LHS = RHS.

Therefore, the statement is proved.

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4 years ago

Given that f (x) = x^2 - 5x and g (x) x + 4

velikii [3]

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3 years ago

What statement best describes inductive reasoning vs deductive reasoning

barxatty [35]

Inductive reasoning works from the root and extends outwards. That is, it starts from the conclusion, then makes premises out of this conclusion. This is opposite to deductive reasoning. The process works from the outside down to the root cause. So, you lay out all the premises first, then deduce the conclusion.

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3 years ago

Please help fastit's for a quiz​

Please help fastit's for a quiz