What is 7 times the sum of a squared and b minus 4 times the sum of a squared and b

1 answer:

7 0

The sum of a squared and b is given by
$a^2+b$

7 times the sum of a squared and b is given by
 $7(a^2+b)$
while 4 times the sum of a squared and b is given by
 $4(a^2+b)$

Therefore, 7 times the sum of a squared and b minus 4 times the sum of a squared and b is given by

$7(a^2+b)-4(a^2+b)=3(a^2+b)$

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Dennis_Churaev [7]

Answer:

~ m∠1 = 35 degrees ( ° ) ~

Step-by-step explanation:

1. This pair of intersecting lines form four pairs of supplementary angles, which may be one approach to this problem.

2. This first pair includes the 145 degree angle, which we may assign as 4, and angle 1, the second being 2 and 1, 2 and 3, and 3 and 145 degrees.

3. If we were to consider the first pair of supplementary angles, it would be that 145 + m∠ 1 = 180, or ⇒ m∠1 = 180 - 145 = 35 degrees ( ° )

4. To confirm that this is the right answer, let us prove that with this measure of ∠1 the intersecting lines = 360 degrees, as after all they form a circle.

5. By Vertical Angles Theorem: m∠2 = m∠4, and m∠3 = m∠1

6. It is provided that m∠4 = 145 degrees ( ° ) and m∠1 = 35 degrees ( ° ). Given such let us substitute these values into Step #5, as such ⇒ m∠2 = 145°, and m∠3 = 35°

7. The sum of the angles are known to 360 degrees, as provided previously, such that m∠1 + m∠2 + m∠3 + m∠4 = 360°, ⇒ 35 + 145 + 35 + 145 = 360, ⇒ 360 = 360

8. This proves that the m∠1 = 35 degrees ( ° )

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