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3 years ago

Subtract: −11−(−6) please help i can noot g iit

Mathematics

2 answers:

nexus9112 [7]3 years ago

5 0

-5 is the answer
-11-(-6)= -5

andriy [413]3 years ago

3 0

Answer:

-5

Step-by-step explanation:

-11-(-6)

-11+6

-5

If this helps please mark as brainliest

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Answer:

Part 1) The length of each side of square AQUA is $3.54\ cm$

Part 2) The area of the shaded region is $(486\pi-648)\ units^{2}$

Step-by-step explanation:

Part 1)

step 1

Find the radius of the circle S

The area of the circle is equal to

$A=\pi r^{2}$

we have

$A=25\pi\ cm^{2}$

substitute in the formula and solve for r

$25\pi=\pi r^{2}$

simplify

$25=r^{2}$

$r=5\ cm$

step 2

Find the length of each side of square SQUA

In the square SQUA

we have that

SQ=QU=UA=AS

$SU=r=5\ cm$

Let

x------> the length side of the square

Applying the Pythagoras Theorem

$5^{2}=x^{2} +x^{2}$

$5^{2}=2x^{2}$

$x^{2}=\frac{25}{2}\\ \\x=\sqrt{\frac{25}{2}}\ cm\\ \\ x=3.54\ cm$

Part 2) we know that

The area of the shaded region is equal to the area of the larger circle minus the area of the square plus the area of the smaller circle

Find the area of the larger circle

The area of the circle is equal to

$A=\pi r^{2}$

we have

$r=AB=18\ units$

substitute in the formula

$A=\pi (18)^{2}=324\pi\ units^{2}$

step 2

Find the length of each side of square BCDE

we have that

$AB=18\ units$

The diagonal DB is equal to

$DB=(2)18=36\ units$

Let

x------> the length side of the square BCDE

Applying the Pythagoras Theorem

$36^{2}=x^{2} +x^{2}$

$1,296=2x^{2}$

$648=x^{2}$

$x=\sqrt{648}\ units$

step 3

Find the area of the square BCDE

The area of the square is

$A=(\sqrt{648})^{2}=648\ units^{2}$

step 4

Find the area of the smaller circle

The area of the circle is equal to

$A=\pi r^{2}$

we have

$r=(\sqrt{648})/2\ units$

substitute in the formula

$A=\pi ((\sqrt{648})/2)^{2}=162\pi\ units^{2}$

step 5

Find the area of the shaded region

$324\pi\ units^{2}-648\ units^{2}+162\pi\ units^{2}=(486\pi-648)\ units^{2}$

7 0

3 years ago