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3 years ago

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Mathematics

1 answer:

qaws [65]3 years ago

8 0

Can you please explain more about this problem?I can’t understand this

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Please help me with this circle area problem!!

kramer

D. 32 square inches.

$ ^ $

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3 years ago

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A square corner of 16 square centimeters is removed

oee [108]

Option C:

Area of the remaining paper = (3x – 4)(3x + 4) square centimeter

Solution:

Area of the square paper = $9x^2$ sq. cm

Area of the square corner removed = 16 sq. cm

Let us find the area of the remaining paper.

Area of the remaining paper = Area of the square paper – Area of the corner

Area of the remaining = $9x^2-16$

= $(3x)^2-(4)^2$

Using algebraic formula: $a^2-b^2=(a+b)(a-b)$

$=(3x-4)(3x+4)$

Area of the remaining paper = (3x – 4)(3x + 4) square centimeter

Hence (3x – 4)(3x + 4) represents area of the remaining paper in square centimeters.

8 0

3 years ago

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1) what is 1/5 times 1/2 ?

Savatey [412]

1. 1/10
2. 3/5
3. 1/10
4. 1/6
5. 3/14

8 0

3 years ago

Squareroot(y-1) + 3 = y

jek_recluse [69]

Note:xy means x times y and x(y) means the same thing

so
first we get rid of square root then
make the equation equal to zero becaues if
xy=0 then x or/and y=0

squareroot(y-1)+3=y

isolate the squareroot
subtrac 3 from boht sides

squareroot(y-1)=y-3

square both sides (since they are equal, you should be able to square both sides and still make it true)

(squareroot(y-1))^2=(y-3)^2
(y-1)=(y-3)(y-3)
y-1=y^2-6y+9
subtrac y from both sides
-1=y^2-7y+9
add 1 to both sides
0=y^2-7y+10
find what two number multiply to make 10 and add to get -7
the answer is -2 and -5

0=(y-5)(y-2)
therfore
y-5=0
and/or
y-2=0

therefor
y=5 or/and 2 might work
let's try out 2
square root(2-1)+3=2
square root(1)+3=2
1+3=2
false
so 2 doesn't work

let's try 5
squareroot(5-1)+3=5
squareroot(4)+3=5
2+3=5
5=5
true

y=5

7 0

3 years ago

Three consecutive integers can be represented by the variable n, n+1 and n+2. If the sum of three consecutive integers is 57, wh

pashok25 [27]

You have some unknown integer $n$ , and you know that adding this and the next two integers, $n+1$ and $n+2$ , gives a total of 57.

This means

$n+(n+1)+(n+2)=57$

The task is to find all three unknown integers. Notice that if you know the value of $n$ , then you pretty much know the value of the other three integers.

To find $n$ , solve the equation above:

$n+(n+1)+(n+2)=3n+3=57\implies 3n=54\implies n=\dfrac{54}3=18$

So if 18 is the first integer, then others must be 19 and 20.

7 0

3 years ago

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