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11Alexandr11 [23.1K]
3 years ago
12

A rectangle has a length of 3.5 inches and a width of 2.5 the perimeter is 6 inches what is the area of a rectangle rounded to t

he nearest tenth?
Mathematics
1 answer:
Alexeev081 [22]3 years ago
5 0

Answer:

8.8

Step-by-step explanation:

A=lw

A= (3.5)(2.5)

A= 8.75 ≅ 8.8

<em>plz mark me brainliest. </em>:)

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What is 5/6 of 11/12 ?
gulaghasi [49]

Answer:

55/72

Step-by-step explanation:

5/6 of 11/12

= 5/6 x 11/12

=55/72

8 0
3 years ago
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Could someone please help me answer this question? I would really appreciate a through explanation, how to work it, the answer t
Shtirlitz [24]
-|-10|-(-10)
A negative outside of the “ | “ means that one number inside will be negative . -10-(-10)
An imaginary | will “appear” in between the two negative signs or it will turn into an addition sign.

-10+10
A positive subtracts negatives from the negative number. The answer is 0

I hope this helped.
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3 years ago
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PLEASE HELP ! worth 18 points :)
leonid [27]

Answer:

x = 6.6

Step-by-step explanation:

law of cosines:

x² = b² + c² - 2ac · cosX

x² = 5² + 8² - 2(5 · 8) · cos 55°

x² = 25 + 64 - 80· cos 55°

x² = 25 + 64 - 45.89

x² = 89 - 45.89

x² = 43.11

x = √43.11

x ≈ 6.6

3 0
3 years ago
K/5.6+ 41.5 = 10.8 <br>The solution is k=​
Contact [7]

Answer: The solution is k = -171.92.

4 0
3 years ago
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Circle O, with center (x, y), passes through the points A(0, 0), B(–3, 0), and C(1, 2). Find the coordinates of the center of th
balandron [24]
The answer:

the main formula of the circle's equation is 
(x-a)²+ (y-b)² = R²
where C(a, b) is the center of the circle
R is the radius

if a point A(x', y') passes through the circle, so the equation of the circle can be written as 
(x'-a)²+ (y'-b)² = R², and that is  a main formula.


<span>Circle O, with center (x, y), passes through the points A(0, 0), B(–3, 0), and C(1, 2), so we have exactly three equation:
</span>
(0-x)² + (0-y)² = R², circle O passes through A
x²+y²= R²
(-3 -x)² + (0-y)² = R², circle O passes through B
(-3 -x)² + (y)² = R²
(1-x)² + (2-y)² = R², circle O passes through A
(1-x)² + (2-y)² = R²

and we know that R= OA = OC= OB, 
OA=R= sqrt( (0-x)² + (0-y)² ) = OB = sqrt((-3 -x)² + (0-y)²), this implies

x²+y² = (-3 -x)² + (0-y)² , it implies x² = 9+ x² + 6x , and then -9/6=x, x= -3/2
and when OA = OC
x²+y² =(1-x)² + (2-y)²  so, x²+y² =1+x²-2x +4+y²-4y, therefore -5= -2x -4y
 -5= -2x -4y, when x = -3 /2   we obtain y = 2

the center is C(-3/2, 2)
7 0
4 years ago
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