What fraction has the greatest value 5/6, 7/12, 7/10

1 answer:

8 0

First you have to make them to a common denominator.
5/6 7/12 7/10
50/60 35/60 42/60
from these the greatest value is 35/60 because 35 parts of 60 is bigger than the rest.
so the answer is 7/12

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If the sinø=7/12, what is cosø?

777dan777 [17]

Answer:

If the sinø=7/12, cosø is $\frac{\sqrt{95} }{12}$

Step-by-step explanation:

We are given sinФ = 7/12

We need to find cosФ

The basic trigonometric functions of right triangle are:

$sin\theta=\frac{opposite}{hypotenuse}$

$cos\theta=\frac{adjacent}{hypotenuse}$

We need values of adjacent and hypotenuse to find cosФ

Using Pythagoras theorem we can find the value of adjacent

$(Hypotenuse)^2= (Opposite)^2+(Adjacent)^2$

We have Hypotenuse= 12 and Opposite = 7 (because $sin\theta=\frac{opposite}{hypotenuse}$ and we are given $sin\theta=\frac{7}{12}$

Inserting values and finding adjacent:

$(Hypotenuse)^2= (Opposite)^2+(Adjacent)^2\\(12)^2=(7)^2+(Adjacent)^2\\144=49+(Adjacent)^2\\144-49=(Adjacent)^2\\95=(Adjacent)^2\\\sqrt{(Adjacent)^2} =\sqrt{95}\\Adjacent=\sqrt{95}$

So, value of Adjacent is $\sqrt{95}$

Now finding cosФ

$cos\theta=\frac{adjacent}{hypotenuse}$

Adjacent = $\sqrt{95}$ , hypotenuse = 12

$cos\theta=\frac{\sqrt{95} }{12}$

So, If the sinø=7/12, cosø is $\frac{\sqrt{95} }{12}$

7 0

3 years ago

The average high temperature each month in a calendar year for a city in South Carolina is represented by the function: f(x)=−x2

LuckyWell [14K]

Answer:

$Average\hspace{3}rate\hspace{3}of\hspace{3}change=-7.7$

Step-by-step explanation:

The average rate of change measures the average rate at which a function changes per unit time in a given interval. In another words, the average rate of change is the relationship between vertical and horizontal changes between two points of the function. So, basically if you are trying to find the average rate of change, you are trying to find the slope.

$Average\hspace{3}rate\hspace{3}of\hspace{3}change=Slope=m=\frac{\Delta y}{\Delta x} =\frac{f(x_2)-f(x_1)}{x_2-x_1}$