MATHEMATICS HIGH SCHOOL

Show that the area of a square A inscribed in a circle with radius r is A=2r square​

Answers

Answer 1
Answer:

Answer:

see below and see image.

Step-by-step explanation:

"inscribed" means the four corners (vertices) of the square are on the circle.

The diagonal of the square is the diameter of the circle. Use special right triangles or pythagorean thm to find the side length of the square in terms of r. Use Area formula for a square:

A = s^2 OR s×s

see image.

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Answers

Answer:

Is that all you need help with or you need help with more?

Step-by-step explanation:

Convert 15 years to seconds

Answers

1 hour = 3600 s
1 day = 24 hours = 24 x 3600 = 86400 seconds

1 day ...... 86400 seconds
365 days ..... t seconds

t= 365 x 86400 /1 = 31536000

So, we have 31536000 seconds in 365 days. => 31536000 seconds = 1 year 

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

1 year ........ 31536000 seconds 
15 years .......... z seconds 

z = 15 x 31536000 / 1 
z= 473040000 

So, 15 years = 473040000 seconds !!!!!! 
Done.

Choose the triangle that seems to be congruent to the given one.AFD =~ ____

AFC
BFC
DFE

Answers

Answer:

The triangle that seems to be congruent to triangle AFD is triangle AFC.

Step-by-step explanation:

As in the give triangle AFD, there is an obtuse angle, i.e angle AFD is obtuse angle.

So, in order to have a congruent triangle, the another triangle must have an obtuse angle.

From the given triangles, only triangle AFC has an obtuse angle and that is angle AFC. And they also share a side, so the chances of congruency is more in this case.

Triangle BFC and Triangle DFE does not have an obtuse angle.
AFD is equivalent to AFC.

They are the same size, shape, length, and angles.

I hoped I helped!

What is the solution of the inequality 9 - x^2 < 0?(a) {x | -3 < x < 3}
(b) {x | x > 3 or x < -3}
(c) {x | x | x > 3}
(d) {x | x < -3}

please show how you got the answer. thank you!

Answers

9    -  x²  <  0
-9               -9    to remove 9 deduct it from both sides
      -  x²  < -9
     ÷ -1      ÷-1    to remove the negative sign, divide both sides by -1
         x²  <  9
       √x²  < √9     to remove the exponent of 2, find the square root of both sides.
         x    < 3     Choice A.

To check:

9 -  x²  <  0
9 - 3²   < 0
9 - 9    < 0
0 < 0
\9-x^2\ \textless \ 0\n3^2-x^2\ \textless \ 0\n(3-x)(3+x)\ \textless \ 0\n\{3-x\ \textless \ 0\ \ \wedge\ \ 3+x\ \textgreater \ 0\}\ \ \ \vee\ \ \ \{3-x\ \textgreater \ 0\ \ \wedge\ \ 3+x\ \textless \ 0\}\n\{-x\ \textless \ -3\ \ \wedge\ \ x\ \textgreater \ -3\}\ \ \ \ \ \ \vee\ \ \ \ \ \{-x\ \textgreater \ -3\ \ \wedge\ \ x\ \textless \ -3\}\n~\ \ \ \{x\ \textgreater \ 3\ \ \ \wedge\ \ \ x\ \textgreater \ -3\}\ \ \ \ \ \ \vee\ \ \ \ \ \ \{x\ \textless \ 3\ \ \ \ \wedge\ \ \ x\ \textless \ -3\}\n~\qquad \qquad \ \ x\ \textgreater \ 3\ \qquad \ \ \ \ \ \vee\ \ \ \ \ \qquad \ x\ \textless \ -3\

The solution is  (b) {x | x > 3 or x < -3}

How do i solve and graph 5x- y \geq 5

Answers

one way is to make it into slope intercept form which is
y=mx+b
m=slope
b=y intercept or where the line will go though the y axis
treat the equal to or greater than sign as equals for now

5x-y=5
subtract y from both sides
5x=5+y
subtract 5
5x-5=y
put the sign back in
5x-5 ≥ y
reverse
y ≤ 5x-5

one way is to subsitute values and get values out so
if x=1 then
y≤5(1)-5
y≤0
(x,y)
(1,0)
(2,5)
(3,10)
(4,15)
(0,-5)

plot the points
then the ≤ sign
since ther is an underline, that means that the points are included so draw a solid line through the points

then figure out which side to shade on
the easy way is, is the point (0,0) a solution?
subsitute
0 ≤ 5(0)-5
0 ≤ -5
false so we shade the section that does not include the point (0,0) in it
the graph would look like this attachment
    (you have to put y alone {Ex: y = 12x + 3 (yis alone on one side)}



5x - y ≥ 5       (subract 5x from each side)
-y ≥ -5x + 5   (divide - or -1 from each side)  Note: when you are dividing or multiplying anything negative the sign changes
y ≤ 5x - 5 <--- answer


As for graphing you take the equation and graph it normally, but with a slight change. since y is less than or equal to 5x - 5, you have to color on the spACE BELOW the graphed line.

see picture:

Answer and maybe an explanation.Thanks in advance.

Answers

inverses

solve for x
replace x with f inverse
replace y with x



so
first minus 2 both sides
y-2=√(2x-3)
square both sides
y^2-4y+4=2x-3
add 3 to both sides
y^2-4y+7=2x
divide both sides y 2
(y^2-4y+7)/2=x

x turns to f inverse and y turns to x


f(x)⁻¹=(x^(2)-4x+7)/(2)


looks like they didn't want to expand the binomial

y=\frac{(x-2)^(2)}+3}{2}
answer si A
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