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slamgirl [31]
3 years ago
10

Can you please help me​

Mathematics
2 answers:
STALIN [3.7K]3 years ago
7 0

Answer:

5. 69 degrees

6. 115 degrees

7. 118 degrees

8. 108 degrees

9. 66 degrees

10. 86 degrees

Step-by-step explanation:

Archy [21]3 years ago
5 0

Step-by-step explanation:

5) angle 5 = 69°

than, angle 10 = 69° ( alternate interior angle )

6) angle 6 = 115° ( alternate interior angle )

7) angle 7 = 118° ( alternate interior angle )

8) angle 11 = angle 4 - 180

→ 72 - 180 = 108°

therefore, angle 11 = 108°

9) angle 14 = 114° - 180

→ angle 14 = 66°

10) angle 12 = 86° ( alternate interior angle)

hope this answer helps you dear...take care!

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The perimeter of a rectangle is 46 in and the diagonal is 17 in. Find the area of the rectangle.
OlgaM077 [116]
P=2*(L+l)=46
L+l=46:2
L+l=23
(L+l)^2=529
(L^2+l^2)+2l*L=529

D= V L^2+l^2= 17
L^2+l^2=289

so 289+2l*L=529
     2l*L=529-289=240
l*L=240:2
A=120 


5 0
3 years ago
How do i find the greatest common factor of two numbers​
melamori03 [73]

Answer:

1)you find the least prime factors of each number

2)Multiply those factors both numbers have in common. If there are no common prime factors, the GCF is 1

5 0
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Trevor is painting a mural on a 1-square-meter section of a wall. He has completed a rectangular part of the mural that is 34 me
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Simplify, 5(2x + 3y) - 4(3x - 5y)
borishaifa [10]

5(2x + 3y) - 4(3x - 5y)

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5 0
4 years ago
Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .
kogti [31]

Answer:

The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

b = \sqrt{1 + 2\cdot a \cdot c} (2)

c is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if (x+y)^{2} = x^{2} + 2\cdot x\cdot y  + y^{2}. From statement, we must fulfill the following identity:

a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}

By Associative and Commutative properties, we can reorganize the expression as follows:

a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2} (1)

Then, we have the following system of equations:

x = a (2)

(b^{2}-1) = 2\cdot x\cdot y (3)

y = c (4)

By (2) and (4) in (3), we have the following expression:

(b^{2} - 1) = 2\cdot a \cdot c

b^{2} = 1 + 2\cdot a \cdot c

b = \sqrt{1 + 2\cdot a\cdot c}

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, a, b, c > 1. If a, b and c are prime numbers, then  2\cdot a\cdot c must be an even composite number, which means that a and c can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, b must be a natural number, which means that:

1 + 2\cdot a\cdot c \ge 4

2\cdot a \cdot c \ge 3

a\cdot c \ge \frac{3}{2}

But the lowest possible product made by two prime numbers is 2^{2} = 4. Hence, a\cdot c \ge 4.

The family of all prime numbers such that a^{2} + b^{2} + c^{2} -1 is a perfect square is represented by the following solution:

a is an arbitrary prime number. (1)

b = \sqrt{1 + 2\cdot a \cdot c} (2)

c is another arbitrary prime number. (3)

Example: a = 2, c = 2

b = \sqrt{1 + 2\cdot (2)\cdot (2)}

b = 3

4 0
3 years ago
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