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Alja [10]
2 years ago
14

Help me find the area of this circle

Mathematics
1 answer:
ch4aika [34]2 years ago
3 0
The area would be approximately 153.94.

The area of a circle is pie multiplied by r^2. 

So you have the radius, which is r. Plug in r. Technically you have two numbers because pie equals approximately 3.14.

A=3.14(7)^2 
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This is urgent! Please answer!
Alja [10]

Answer:

hold up im solving it

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Solve the systems of equation by graphing (Picture provided)
11Alexandr11 [23.1K]

Answer:

Option b

Step-by-step explanation:

The following system of linear equations is shown

x + y = 8\\2x + y = 3

These are two different slope lines.

We find the cut points of both lines with the axes.

x + y = 8

Cut with the x axis. (y = 0)

x = 8

Cut with the y axis. (x = 0)

y = 8

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

2x + y = 3

Cut with the x axis. (y = 0)

2x = 3\\x = 1.5

Cut with the y axis. (x = 0)

y = 3

The solution to this system will be a point for which it is fulfilled that:

x + y - 8 = 2x + y-3

In the image, different graphs with intersections are shown.

Locate among the options, one that shows the two lines of the system of equations according to their intersections with the x and y axes.

Option b is the only one that shows the graph of the lines

x + y = 8\\2x + y = 3

Then, The point of intersection of both lines in the graph is:

(-5, 13)

Therefore the solution of the system of equations is: (-5, 13)

You can verify this by replacing the point in the relationship

x + y - 8 = 2x + y-3\\\\(-5) +13 -8 = 2 (-5) +13 -3\\\\0 = 0

Equality is satisfied

Finally the answer is the option b

8 0
3 years ago
Numerical value of cosh(ln5)
strojnjashka [21]
\cosh x=\dfrac{e^x+e^{-x}}2

So

\cosh(\ln 5)=\dfrac{e^{\ln5}+e^{-\ln5}}2=\dfrac{5+\frac15}2=\dfrac{13}5
6 0
3 years ago
The equation of a circle is x2 + y2 + 6x + 4y + 10 = 1. What is this equation written in its standard form?
Over [174]
You need to complete the square in x and in y,

x^{2}  + y^{2} + 6x + 4y + 10 = 1

First, group the x-terms and the y-terms separately.

x^{2} + 6x  + y^{2} + 4y + 10 = 1

Move the 10 to the right side by subtracting 10 from both sides.

x^{2} + 6x  + y^{2} + 4y = -9

Now complete the square in x and in y.
The constant you need to add to complete each square is the square of half of the coefficient of the x or y term. Make sure to add the constants to both sides of the equation.

x^{2} + 6x  + 9 + y^{2} + 4y + 4= -9 + 9 + 4

(x + 3)^{2} + (y + 2)^{2} = 4
8 0
3 years ago
Consider the diagram.
Sergio [31]
I don’t see the image can u re- upload it
6 0
2 years ago
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