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2 years ago

Perform the indicated operation. (y4 - 1) ÷ (y + 1)

2 answers:

7 0

$The\ domain:\\D:y\neq-1\\\\(y^4-1)\div(y+1)=\dfrac{(y^2)^2-1^2}{y+1}=\dfrac{(y^2-1)(y^2+1)}{y+1}\\\\=\dfrac{(y-1)(y+1)(y^2+1)}{y+1}=(y+1)(y^2+1)=y^3+y^2+y+1\\\\Used:\\(a^n)^m=a^{nm}\\\\a^2-b^2=(a-b)(a+b)$

Dmitry_Shevchenko [17]2 years ago

3 0

Answer: y^3 - y^2 + y - 1

Step-by-step explanation: Factor the numerator and denominator and cancel the common factors. This should leave you with y^3-y^2+y-1.

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A two pound box of fruit snacks contains 24 packets. Find the unit rate in packets per pound

Monica [59]

24/2=12

The unit rate is 12 packets per pound.

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1 year ago

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What are the coordinates of the circumcenter of this triangle? Enter your answer in the boxes.

Juliette [100K]

Answer:

(2,1)

Step-by-step explanation:

We can see that the given triangle.

The coordinates of A are (-1,5) .

The coordinates of B are (-1,-3).

The coordinates of C are (5,-3).

Distance formula: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

AB= $\sqrt{(-3-5)^2+(-1+1)^2}=8$ units

BC= $\sqrt{(5+1)^2+(-3+3)^2}=6$ units

AC= $\sqrt{(5+1)^2+(-3-5)^2}=10$ units

Pythagoras theorem: $(Hypotenuse)^2=(Base)^2+(Perpendicular\;side)^2}$

$AB^2+BC^2=8^2+6^2=100$

$AC^2=(10)^2=100$

Therefore, $AC^2=AB^2+BC^2$

When a triangle satisfied the Pythagoras theorem then, the triangle is right triangle.

Hence, the given triangle is a right triangle.

We know that circum-center of right triangle is the mid point of hypotenuse.

Mid-point formula: $x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$

Using this formula then, we get

Mid-point of hypotenuse AC is given by

$x=\frac{-1+5}{2}=2,y=\frac{5-3}{2}=1$

Hence, the circum-center of triangle is (2,1).

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2 years ago

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2 years ago

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Nuetrik [128]

C. The way the sample was chosen may overrepresent or underrepresent students taking certain language classes.

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1 year ago

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denis-greek [22]

Given:

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