The other 2 angles of given right angles are 61.93° and 28.072°, if a triangle with side lengths 8, 15, and 17 is a right triangle by the converse of the Pythagorean Theorem.
Step-by-step explanation:
The given is,
Right angled triangle,
Side lengths are 8, 15, and 17
Step:1
The given triangle is right angle triangle by the converse of Pythagorean theorem, so the trigonometric ratio,
Ref the attachment,
For angle a,
...................................................(1)
Where, Opp - 8
Hyp - 17
From equation (1),
= 0.470588
(0.470588)
a = 28.072°
For angle b,
...................................................(1)
Where, Opp - 15
Hyp - 17
From equation (1),
= 0.882352
(0.882352)
b = 61.93°
Step:2
Check for solution for right angle triangle,
90 ° = Other 2 angles
90 ° = a + b
90 ° = 28.072° + 61.93°
90 ° = 90 °
Result:
The other 2 angles of given right angles are 61.93° and 28.07°, if a triangle with side lengths 8, 15, and 17 is a right triangle by the converse of the Pythagorean Theorem.
12/6 is 2
2 +2 is 4
The overall answer is 4
Answer:
(- 2, - 1 )
Step-by-step explanation:
Given the 2 equations
2x - 3y = - 1 → (1)
x + 4y = - 6 → (2)
Rearrange (2) expressing x in terms of y by subtracting 4y from both sides
x = - 6 - 4y → (3)
Substitute x = - 6 - 4y into (1)
2(- 6 - 4y) - 3y = - 1 ← distribute and simplify left side
- 12 - 8y - 3y = - 1
- 12 - 11y = - 1 ( add 12 to both sides )
- 11y = 11 ( divide both sides by - 11 )
y = - 1
Substitute y = - 1 into (3) for corresponding value of x
x = - 6 - 4(- 1) = - 6 + 4 = - 2
Solution is (- 2, - 1 )
7x² = 9 + x Subtract x from both sides
7x² - x = 9 Subtract 9 from both sides
7x² - x - 9 = 0 Use the Quadratic Formula
a = 7 , b = -1 , c = -9
x =

Plug in the a, b, and c values
x =

Cancel out the double negative
x =

Square -1
x =

Multiply 7 and -9
x =

Multiply -4 and -63
x =

Multiply 2 and 7
x =

Add 1 and 252
x =

Split up the

x =

The approximate square root of 253 is <span>15.905973.
</span>x ≈

Add and subtract
x ≈

Divide
x ≈

Round to the nearest hundredth
x ≈

<span>
</span>
Assignment: 
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Answer: 
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Explanation: 
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[ Step One ] Rewrite 

[ Step Two ] Rewrite Equation

[ Step Three ] Apply Exponent Rule
Note: 

[ Step Four ] Refine

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