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boyakko [2]
3 years ago
9

Please help answer these questions. My teacher said they were really easy but I just don't understand. Will mark brainliest !!!

Mathematics
1 answer:
kodGreya [7K]3 years ago
6 0

Answer:

1.  A = 59

2.  A = 43

Step-by-step explanation:

If we have a right triangle  we can use sin, cos and tan.

sin = opp/ hypotenuse

cos= adjacent/ hypotenuse

tan = opposite/ adjacent


For the first problem, we know the opposite and adjacent sides to angle A

tan A = opposite/ adjacent

tan A = 8.8 / 5.2

Take the inverse of each side

tan ^-1 tan A = tan ^-1 (8.8/5.2)

A = 59.42077313

To the nearest degree

A = 59 degrees


For the second problem, we know the  adjacent side and the hypotenuse to angle A

cos A = adjacent/hypotenuse

cos A = 15.3/21

Take the inverse of each side

cos ^-1 cos A = cos ^-1 (15.3/21)

A = 43.23323481

To the nearest degree

A = 43 degrees


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If EG = 2u + 9 and EI = 5u - 48, find EI in parallelogram FGHI.
zalisa [80]

Answer:

47

Step-by-step explanation:

The diagonals of a parallelogram bisect in their middle point, so :

2u + 9 = 5u - 48

3u = 57

u = 19

EI = 5* 19 - 48 = 47

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3 years ago
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Answer:

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Is that correct?

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Which equation represents a line which is perpendicular to the line y=-4/5x+2?
lyudmila [28]

Answer:

D) 4x + 5y = -30 = -4/5

Step-by-step explanation:

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<span>x^2+4x-1/(x+2)^2=0
x^2 + 4x - 1 = 0
</span>
\frac{x^2+10x-7}{x^2+10x+25} =0\\x^2+10x-7=0\\x= \frac{-b\pm\sqrt{b^2-4ac}}{2a};
where a = 1, b = 4 and c = -1

x=\frac{-4\pm\sqrt{4^2-(4\times1\times(-1)}}{2\times1}\\=\frac{-4\pm\sqrt{16+4}}{2}=\frac{-4\pm\sqrt{20}}{2}\\=\frac{-4\pm\sqrt{4\times5}}{2}=\frac{-4\pm2\sqrt{5}}{2}=-2\pm\sqrt{5}\\=-2+\sqrt{5}&#10; \ or \ -2-\sqrt{5}\\x=0.24 \ or \ x = -4.24
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