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3 years ago

14

Is 3z-1 in standard form?

1 answer:

babunello [35]3 years ago

7 0

Answer:

Yes because

Step-by-step explanation:

Is 3z-1 in standard form?

Yes becausestandard form of a polynomial, the value in the power of the variable such as z in the equation reduces from left to right. Therefore, 3z - 1 is in standard form since the z term comes to the left of the constant term

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4 groups of ten

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175-135=40

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3 years ago

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3 years ago

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3 years ago

State whether the following coordinates on a Cartesian plane form an acute, obtuse or right triangle: a) (-1, 1), (7,-2) and (1,

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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