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Andrej [43]
3 years ago
11

Please help me asap!​

Mathematics
1 answer:
zysi [14]3 years ago
7 0

Answer:

Let's analyze the example first:

We have:

3, 4, 6      target 42    answer: (3 + 4)*6

If we solve the equation, we get:

(3 + 4)*6 = 7*6 = 42

So, we need to reorder the first 3 numbers in an expression, such that the solution of the expression is the target number.

a) 2, 5, 6     Target 40

We know that 5*8 = 40

and that 2 + 6 = 8

Then the answer is:

(2 + 6)*5

b) 3, 4, 6  target 12

Here we know that 4*3 = 12

and 6 - 3 = 3

Then the answer is:

(6 - 3)*4

c) 4, 6, 10   target 1.

We know that the quotient of two equal numbers is 1.

And 4 + 6 = 10

Then the answer here is:

(4 + 6)/10

d) 3, 5, 6 target 21

We know that 3*5 = 15

And 15 + 6 = 21

Then the answer here is:

3*5 + 6

e) 2, 7, 10  target 6.

Here is a harder one.

We know that 2*3 = 6

And that 10 - 7 = 3

Then the answer ere is:

2*(10 - 7)

f) 3, 6, 15   target  3

Here we can use:

(15 - 6) = 9

And 9/3 = 3

Then the answer is:

(15 - 6)/3

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Step-by-step explanation:

Given equation:

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\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

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

Using the <u>Zero Product Property</u>, 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}

Therefore, the solutions to the given equation are: x = 1 or x = -1

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