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Anna [14]
4 years ago
12

Solve the expression using the order of operation: Find the hypotenuse of a right triangle if the legs are 8cm and 5cm by using

the Pythagorean theorem: (leg^2 +leg^2=hypotenuse^2)
Round your answer to the nearest hundredth if necessary
Hypotenuse =. Cm


I got 89^2 but I don’t think I I did it correctly
Mathematics
1 answer:
DochEvi [55]4 years ago
4 0

{a}^{2} +  {b}^{2} =  {c}^{2}  \\  {8}^{2} +  {5}^{2} =  {c}^{2}  \\ 64 + 25 =  {c}^{2}  \\ 89 =  {c}^{2}  \\  \sqrt{89} =  {c}^{2}  \\ 9.4 = c
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How would I solve this? i know how to solve easier variations but not like this the question is:
IRISSAK [1]

The linear function with values of f(-1)=8 and f(5)=6 is  f(x) = \frac{-x+23}{3}.

What is linear function?

 A straight line on the graph has the equation or formula y = f(x) = px + q is linear function .One independent and one dependent variables are present. X and Y are the independent and dependent variables, respectively. P is the y-intercept or constant term, and it also represents the value of the dependent variable. A straight line on the coordinate plane is represented by a linear function.

Here the given  values are f(-1)=8 and f(5)=6.

We know that f(x)=y , Then the points are (-1,8) and (5,6).

Using slope formula ,

=> slope m= \frac{y_2-y_1}{x_2-x_1}

=> slope m = -2/6=-1/3

Now using slope formula,

=> y-y_1=m(x-x_1)

=>y-8=-1/3(x+1)

=>3y-24=-x-1

=> 3y=-x+23

=> y = \frac{-x+23}{3}

Hence the linear function is f(x) = \frac{-x+23}{3}.

To learn more about linear function refer the below link

brainly.com/question/28899900

#SPJ9

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1 year ago
Carly is a catcher on the school softball team. By the end of the season, she had played 45% of the 49 games the team played. In
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Step-by-step explanation:

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The equation of a line is y=3. Write an equation in slope intercept form of a line parallel to y=3 that passes through (0,6).
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Answer:

y = 6

Step-by-step explanation:

We have to consider the equation of a line ( in slope - intercept form ) that is parallel to y = 3, and passes through the point ( 0, 6 ). Now y = 3 can be plotted as a horizontal line that passes through the point ( 0, 3 ). If we want a line that is parallel to this, it must be horizontal as well;

Let us consider the second point now. This line must pass through the point ( 0, 6 ). We can conclude that the line must be 1. Horizontal, and 2. Pass through point ( 0, 6 );

Equation - y = 6

7 0
3 years ago
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