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Jlenok [28]
3 years ago
5

What is the awnser [(12−5)⋅12]÷6

Mathematics
2 answers:
prohojiy [21]3 years ago
6 0
We Need to First Do 12 - 5.
12 - 5 = 7.
[7 * 12] <span>÷ 6
12 * 7 = 84.
84 </span><span>÷ 6 
This Is Our Answer:
14.</span>
masya89 [10]3 years ago
6 0
12-5=7.12/6=1.1866667
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How to represent 0.5 on a decimal grid
Alina [70]

Answer:

put one and then put 0.5 in between 0 an 1

Step-by-step explanation:

7 0
2 years ago
Hey plsss help meeeeee
nalin [4]

Answer:

The first, third, and fourth answer choices represent a function.

Step-by-step explanation:

A relation is a relationship between sets of values. The two quantities that are being related to each other are the input (x-variable) and the output (y-variable). But relations in general aren't always a good way to relate between x and y.

Say that I have situation where I want to give <em>x </em>dollars to the cashier so he can change them to <em>y</em> quarters. Here is a "example" of the relation:

Dollars (x) | Quarters (y)            

----------------------------------              

       0       |          0                      

        1       |          4                      

       2       |          8

       2       |          12

Do you see something wrong here? Yes! We all know that you can't exchange 2 dollars for 12 quarters. You can only exchange 2 dollars for 8 quarters and only 8 quarters. This is a general reason why we don't rely on general relations for real-life situations. One x-variable does not exactly map to one and only one y-variable.

However, a relation that can map one x-variable to one and only one y-variable is known as a function. Let's make the above example an actual function to prove a point:

Dollars (x) | Quarters (y)            

----------------------------------              

       0       |          0                      

        1       |          4                      

       2       |          8

       3       |          12

Now, the 3 dollars make 12 quarters as it should. This is how a function should look like.

There are two ways to check if a relation is a function. On a relation, table, or a set of ordered pairs, you have to make sure there is no "x-variable" that repeats. All x-values of a relation have to be unique in order to be a function. On a graph, you can also perform the Vertical Line Test. If you draw vertical lines over a relation and if the lines cross only once, then it is a function. If not, it fails the Vertical Line Test.

So to answer you're question, the first, third, and fourth choices are functions because they all have unique x-variables. The second choice is not a function because it fails the Vertical Line Test.

7 0
2 years ago
In the first round of a card game, niki scored 20 points. Then, she lost 10 points on one turn and a additional 25 points on her
Furkat [3]
20-10= 10 + 25 = 35!!!
5 0
3 years ago
PLS HELP ME ASAP GUYS!! THANK U SM!! HELP! Thank u!!
Naya [18.7K]
Asked and answered previously.
brainly.com/question/8726021
4 0
2 years ago
Find the taylor polynomial t3(x) for the function f centered at the number
inysia [295]

Answer:

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

Step-by-step explanation:

We are given that

f(x)=7tan^{-1}(x)

a=1

T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}

Substitute n=3 and a=1

t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}

f(x)=7tan^{-1}(x)

f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}

Where tan^{-1}(1)=\frac{\pi}{4}

f'(x)=\frac{7}{1+x^2}

Using the formula

\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}

f'(1)=\frac{7}{2}

f''(x)=\frac{-14x}{(1+x^2)^2}

f''(1)=-\frac{7}{2}

f''(x)=-14x(x^2+1)^{-2}

f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})

By using the formula

(uv)'=u'v+v'u

f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}

f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}

f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}

Substitute the values

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

7 0
3 years ago
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