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svetoff [14.1K]
3 years ago
11

Anna and Alice share a rectangle bedroom that is 18 feet long and 14 feet wide. they split the room into two halves by placing a

piece of tape from the northwest corner of the room to the southeast corner of the room. what is the area of each girls half of the room?
Mathematics
2 answers:
Alja [10]3 years ago
4 0
The area of the whole room = 14 × 18 = 252

Since they divide it by half, divide the are by 2.

252 ÷ 2 = 126

Answer = 126 ft²

Hope this helped☺☺
Marat540 [252]3 years ago
3 0
126 is the area of one hafe of the room
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Please help me with this. I think I understand it and know what it is but I just want someone to check it for me to see if I'm r
arsen [322]
Ok, so
Length = L
Width = L + 7

Since area = length x width, then

30 = L × (L + 7)

Remember that L + 7 Is width so...

L + 7 = 30/L

So the expression for the width in terms of length would be...
w = 30/L - 7
8 0
3 years ago
Does anyone know the answer to this question
lys-0071 [83]

ANSWER

\cos B = \frac{ \sqrt{3} }{3}

EXPLANATION

The given triangle is a right triangle.

It was given that,

a = 1

and

b =  \sqrt{2}

Using the Pythagoras Theorem, we can determine the value of c.

{c}^{2}  =  {( \sqrt{2} )}^{2}  +  {1}^{2}

{c}^{2} = 2 +  1

{c}^{2} = 3

c =  \sqrt{3}

The ratio is the adjacent over the hypotenuse.

\cos B =  \frac{1 }{ \sqrt{3} }

We rationalize to get:

\cos B =  \frac{ \sqrt{3}  }{ \sqrt{3}  \times  \sqrt{3} }  =  \frac{ \sqrt{3} }{3}

3 0
3 years ago
Consider the following sets of matrices: M2(R) is the set of all 2 x 2 real matrices; GL2(R) is the subset of M2(R) with non-zer
defon

Answer:

Step-by-step explanation:

REcall the following definition of induced operation.

Let * be a binary operation over a set S and H a subset of S. If for every a,b elements in H it happens that a*b is also in H, then the binary operation that is obtained by restricting * to H is called the induced operation.

So, according to this definition, we must show that given two matrices of the specific subset, the product is also in the subset.

For this problem, recall this property of the determinant. Given A,B matrices in Mn(R) then det(AB) = det(A)*det(B).

Case SL2(R):

Let A,B matrices in SL2(R). Then, det(A) and det(B) is different from zero. So

\text{det(AB)} = \text{det}(A)\text{det}(B)\neq 0.

So AB is also in SL2(R).

Case GL2(R):

Let A,B matrices in GL2(R). Then, det(A)= det(B)=1 is different from zero. So

\text{det(AB)} = \text{det}(A)\text{det}(B)=1\cdot 1 = 1.

So AB is also in GL2(R).

With these, we have proved that the matrix multiplication over SL2(R) and GL2(R) is an induced operation from the matrix multiplication over M2(R).

7 0
3 years ago
Each of the 25 students in Mr Walker's class sold 16 raffle tickets. If each ticket costs $15 how much money did Mr Walker's stu
ludmilkaskok [199]

Answer: $6000

Hope this helps!-xoxo

5 0
3 years ago
Which of the following is NOT a factor of b3 +6b²?<br> 63<br> 062<br> Ob<br> (b + 6)
Gnesinka [82]

Answer:

b³

Step-by-step explanation:

The lowest common multiple of the expression is b². This because, taking out a factor of b, b ( b² + 6b ), taking out a factor of b², b² ( b + 6 ),

taking out a factor of ( b + 6 ), ( b + 6 ) b². As a result b² is only the option which is not a factor.

7 0
2 years ago
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