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3 years ago

Helpp i dont want to fail math

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

6 0

Answer:

54.7 pounds left

Step-by-step explanation:

set up the equation

1015-(9(39.2)+25(15.5)+40(5.5)

simplify the parenthesis

1015-(352.8+387.5+220)

1015-960.3

54.7

hope this helps UwU

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3 years ago

PLEASE HELP WHEN YOU SEE THIS

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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3 years ago

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anzhelika [568]

Answer:

AB ≈ 14.3

Step-by-step explanation:

We're given two sides (BC and CA) and an angle (C) between them; the law of cosines is a good tool for calculating the third side of the triangle here. To remind you, the law of cosines tells us the relationship between the sides of a triangle with side lengths a, b, and c:

$c^2=a^2+b^2-2ab\cos{C}$

Where C is the angle between sides a and b. c is typically the side we're trying to find, so on our triangle, we have

$c=AB\\a=BC=16\\b=CA=5\\C=m\angle C=61^{\circ}$

Substituting these values into the law of cosines:

$c^2=16^2+5^2-2(16)(5)\cos{61^{\circ}}\\c^2=256+25-160\cos{61^{\circ}}\\c^2=281-160\cos{61^{\circ}}\\c=\sqrt{281-160\cos{61^{\circ}}}\\c\approx 14.3$

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