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astra-53 [7]
3 years ago
10

How Do You Began The Dividing polynomial

Mathematics
1 answer:
Minchanka [31]3 years ago
7 0

Answer:

y^3 - 7y^2 - 16y +64

Step-by-step explanation:

start by distributing y and - 4 to the equation to get ...

y^3 - 8y^2 + 16y = -y^2 + 32y - 64

then cross factor the number with the same exponents to get ...

y^3 - 7y^2 - 16y + 64

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Draw a line representing the "run" and a line representing the "rise" of the line. State the slope of the line in simplest form.
AnnyKZ [126]

Answer:

  -5/4

Step-by-step explanation:

See the attachment for rise and run lines. The slope is the ratio ...

   slope = rise / run = -5/4

4 0
3 years ago
40 points Please help!!!
pickupchik [31]

Answer:

V = 24.28 in ^3

Step-by-step explanation:

The area of the base is

A =5/2 × s × a  where s is the side length and a is the apothem

A = 5/2 ( 2.13) * .87

A = 4.63275

The volume is

V = Bh  where B is the area of the base and h is the height

V = 4.63275 ( 5.24)

V =24.27561 in^3

Rounding to the  hundredth

V = 24.28 in ^3

5 0
3 years ago
PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU B
Vesnalui [34]

Answer:

(4,0)

Step-by-step explanation:

5 0
3 years ago
Help!!!!!!!!!!!!!!!!​
kaheart [24]

Answer:

\cos\alpha\sin\alpha(\cos\alpha-\sin\alpha)

Step-by-step explanation:

First, simplify each term:

\sin\left(\dfrac{\pi}{2}+\alpha\right)=\cos \alpha\\ \\\cos \left(\dfrac{\pi}{2}+\alpha\right)=-\sin \alpha\\ \\\cos \left(\alpha-\dfrac{3\pi}{2}\right)=-\sin \alpha\\ \\\sin \left(\dfrac{3\pi}{2}+\alpha\right)=-\cos \alpha

Then given expression is equivalent to

\cos ^3\alpha+(-\sin \alpha)^3-(-\sin \alpha)+(-\cos \alpha)\\ \\=\cos ^3\alpha-\sin^3 \alpha+\sin \alpha-\cos \alpha\\ \\=(\cos\alpha-\sin\alpha)(\cos^2\alpha+\cos\alpha\sin\alpha+\sin^2\alpha)-(\cos\alpha-\sin\alpha)\\ \\=(\cos\alpha-\sin\alpha)(1+\cos\alpha\sin\alpha-1)\ \ [\cos^2\alpha+\sin^2\alpha=1]\\ \\=\cos\alpha\sin\alpha(\cos\alpha-\sin\alpha)

6 0
3 years ago
Consider the line
AveGali [126]
Y - y₁ = m(x - x₁)
y - 5 = ⁵/₆[x - (-6)]
y - 5 = ⁵/₆(x + 6)
y - 5 = ⁵/₆(x) + ⁵/₆(6)
y - 5 = ⁵/₆x + 5
  + 5          + 5
     y = ⁵/₆x + 10

y - y₁ = m(x - x₁)
y - 5 = -1¹/₅[x - (-6)]
y - 5 = 1¹/₅(x + 6)
y - 5 = 1¹/₅(x) + 1¹/₅(6)
y - 5 = 1¹/₅x + 7¹/₅
  + 5           + 5
     y = 1¹/₅x + 12¹/₅
8 0
3 years ago
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