What is the solution to

Mathematics

1 answer:

miv72 [106K]3 years ago

3 0

For this case, what you must do is to observe in the table the values that take both functions at the moment of being evaluated with different values of x.
We have then that:
for x = 0
f (0) = 1
g (0) = 1
For x = 1
f (1) = 3
g (1) = 3
Therefore, in both cases:
f (x) = g (x)
Thus, the solution values are:
C. x = 0
D. x = 1
answer:
C. x = 0
D. x = 1

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Step-by-step explanation:

Hi student, let me help you out!

....................................................................................................................................

Let's use the acronym PEMDAS. With the help of this little acronym, we will not make mistakes in the Order of Operations! :)

$\dag\textsf{Acronym \: PEMDAS}$

P=Parentheses,

E=Exponents,

M=Multiplication,

D=Division,

A=Addition,

S=Subtraction.

Now let's start evaluating our expression, which is $\mathsf{6-(\cfrac{2}{3})^2}$

According to PEMDAS, the operation that we should perform is "E-Exponents".

Notice that we have a fraction raised to a power. When this happens, we raise both the numerator (2 in this case) and the denominator (3 in this case) to that power, which is 2. After this we obtain $\mathsf{6-\cfrac{4}{9}}$ .

See, we raised both the numerator and the denominator to the power of 2.

Now what we should do is subtract fractions.

Note that 6 and -4/9 have unlike denominators. First, let's write 6 as a fraction: $\mathrm{\cfrac{6}{1}-\cfrac{4}{9}}$ . Now let's multiply the denominator and the numerator of the first fraction times 9: $\mathrm{\cfrac{54}{9}-\cfrac{4}{9}}$ .

See, now the fractions have the same denominator. All we should do now is subtract the numerators: $\mathrm{\cfrac{50}{9}}$ .