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3 years ago

Please help, due tomorrow, please do all!

Mathematics

1 answer:

zlopas [31]3 years ago

4 0

Answer:

are we supposed to multiply

Step-by-step explanation:

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Given that the product of 22 integers is equal to 1, prove that their sum is not equal to 0.

Kay [80]

1. If the product of these integers is to be 1, then all of them must be either 1 or -1.

2. Since the product is positive (+1), it must be that there are an *even* number of negative ones (-1), if any.

3. If the sum were 0 it would mean that the number of +1's must equal the number of -1's. So that means there would have to be exactly 22/2=11 of each.

4. But if there were 11 of each, that means the number of -1's would be *odd* and there's no way the product could be +1 (as stated in 2 above).

Hence, the sum is never 0, if the product of 22 integers is equal +1.

4 0

3 years ago

Read 2 more answers

For the function f(x) = x^2+ 3, what are the outputs for the inputs -3, 2, 0, and 5?

AysviL [449]

Answer:

F(-3)=12

F(2)=7

F(0)=3

F(5)=28

Step-by-step explanation:

8 0

4 years ago

Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

Zero Product Property: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the Zero Product Property, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$