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Ket [755]
3 years ago
13

What are the known solutions to f(x)=g(x)

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer:

<em>1 and 9</em>

Step-by-step explanation:

<u>Common Solutions to Functions</u>

Let f(x) and g(x) be any type of function. When we need to find values of x such that f(x)=g(x), we usually set up the equation for each of them and solve it for x by any available algebraic or approximate method.

In this situation, we are not given the explicit expression of any of the functions, but the table of values for each one of them. We have to take a look of the values of x for which f and g are the same.

A closer look at the table reveals that f(x)=g(x) for two values of x:

for x=1, f(1) = 14 , g(1) = 14

for x = 9, f(9) = 6, g(9) = 6

The answers are 1 and 9

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5. The temperature was
Vesna [10]

Answer:

6.2 is the difference

Step-by-step explanation:

4 0
3 years ago
Explain how modeling partial products can be used to find the products of greater numbers ?
liq [111]

You can break large numbers into a sum of a multiple(s) of 10 and the last digit of the number. For example, you can break 26 as 20+6, or 157 as 100+50+7.

Then, using the distributive property, you can turn the original multiplication into a sum of easier multiplications. For example, suppose we want to multiply 26 and 37. This is quite challenging to do in your mind, but you can break the numbers as we said above:

26\times 37=(20+6)(30+7) = 20\times 30+20\times 7+30\times 6+6\times 7

All these multiplications are rather easy, because they either involve multiples of 10 of single-digit numbers:

20\times 30+20\times 7+30\times 6+6\times 7 = 600+140+180+42=962

3 0
3 years ago
1- Which comparison is NOT true?
Irina18 [472]

Answer:

D

Step-by-step explanation:

Process of Elimination:

A: 3.976 is greater than 3.97

B: 1.025 is lesser than 1.25

C: 9.145 is greater than 9.14

All of the options checked are true, leaving the answer only to be D.

6 0
2 years ago
If k‐th term of a sequence is ak = ( -1) ^k(2-k)k/2k-1
max2010maxim [7]

Given:

kth term of a sequence is

a_k=\dfrac{(-1)^k(2-k)k}{2k-1}

To find:

The next term a_{k+1} when k is even.

Solution:

We have,

a_k=\dfrac{(-1)^k(2-k)k}{2k-1}

Put k=k+1, to get the next term.

a_{k+1}=\dfrac{(-1)^{k+1}(2-(k+1))(k+1)}{2(k+1)-1}

If k is even, then k+1 must be odd and odd power of -1 gives -1.

a_{k+1}=\dfrac{(-1)(2-k-1)(k+1)}{2k+2-1}

a_{k+1}=\dfrac{(-1)(1-k)(1+k)}{2k+1}

a_{k+1}=-\dfrac{1^2-k^2}{2k+1}    [\because (a-b)(a+b)=a^2-b^2]

a_{k+1}=-\dfrac{1-k^2}{2k+1}

Therefore, the next term is a_{k+1}=-\dfrac{1-k^2}{2k+1}.

7 0
3 years ago
PLEASE HELP!!!!! IT IS URGENT
Marat540 [252]

Answer:

Step-by-step explanation:

B:  to get 4/8 = 1/2 That is not the whole answer, but it gets one started.

D: Very large because -4 -(-9) = 5 That's the power. This is going to make the result large.

That's all I would pick, but the second from the bottom might be an answer.

4 0
3 years ago
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