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avanturin [10]
3 years ago
14

PLEASE HELP ASAP WILL GIVE 30 POINTS

Mathematics
1 answer:
kykrilka [37]3 years ago
3 0

Answer:

C. \displaystyle (−∞, 5) ∪ (5, ∞)

Step-by-step explanation:

This is obvious the correct answer, considering that the denominator can NEVER equal zero.

I am joyous to assist you anytime.

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Rewrite the equation by completing the square.<br> x^2 + 7x + 12 = 0
german

Answer:

x^2 + 7x + 12 = 0

x^2 + 7x  = -12

(+3)(+4)=0

=−3

=−4

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Hope This Helps!!!

3 0
3 years ago
Read 2 more answers
State whether the following coordinates on a Cartesian plane form an acute, obtuse or right triangle: a) (-1, 1), (7,-2) and (1,
LiRa [457]

Answer:

a) Acute triangle

b) Right triangle

Step-by-step explanation:

∵ A triangle having sides a, b and c is called,

Acute : If a² + b² > c² or a² + c² > b² or b² + c² > a²,

Obtuse : if a² + b² < c² or a² + c² < b² or b² + c² < a²

Right : a² + b² = c² or a² + c² = b² or b² + c² = a²,

a) Let A≡(-1, 1), B≡(7,-2) and C≡(1,-5),

By the distance formula,

AB=\sqrt{(7-(-1))^2+(-2-1)^2}=\sqrt{(7+1)^2+(-3)^2}=\sqrt{8^2+3^2}=\sqrt{64+9}=\sqrt{73}\text{ unit}

Similarly,

BC=\sqrt{45}\text{ unit}

CA=\sqrt{40}\text{ unit}

∵ The sum of any two sides is greater than third side,

So, ABC is a triangle,

Now,

AB^2 + BC^2 > CA^2

⇒ ABC is an acute triangle.

b) Let P≡(0,6), Q≡(1,2) and R≡(5,3),

By the distance formula,

PQ = √17 unit,

QR = √17 unit,

RP = √34 unit,

∵ The sum of any two sides of PQR is greater than third side,

⇒ PQR is a triangle ,

Also,

RP^2=PQ^2+QR^2

Hence, PQR is a right triangle.

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3 years ago
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Gala2k [10]
7 times 7 times 7 times 7 times 7 times 7 times 7 times 7 times 7 times 7

5 0
2 years ago
Sin(10°)=?<br>how to solve this??
Triss [41]
Since 10 is going to the 0 power, the answer should be 1. Anything to the zero power is one
7 0
3 years ago
Please help!! Giving BRAINLIEST!
garri49 [273]
Answer: D
Explanation: rad 36 is the same as rad 6^2 and they are perfect roots do then you just add the numbers like usual.
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2 years ago
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