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2 years ago

Write a polynomial function of least degree with rational coefficients so that P(x)=0 has the given roots.

Mathematics

1 answer:

Fittoniya [83]2 years ago

3 0

Answer:

If P(x) has the roots -11 and 6, then its multipliers can be written as

Step-by-step explanation:

P(x)=(x+11)(x-6)

P(x)=x2-6x+11x-66

P(x) = x2+5x-66

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The length of a rectangle is twice that of its width. If the area of the rectangle is 72 square inches, then find the length and

Blizzard [7]

Well, we should start by listing the factors

1,2,3,4,6,8,9,12,16,24,36,72

You can put these in pairs if it will help you,

1*72 2*36 3*24 4*16 6*12 8*9

If you look, you can see that in only one pair the length is twice the width.

So your answer is

W=6

L=12

6 0

2 years ago

Find AB. Round to the nearest tenth. 61°

anzhelika [568]

Answer:

AB ≈ 14.3

Step-by-step explanation:

We're given two sides (BC and CA) and an angle (C) between them; the law of cosines is a good tool for calculating the third side of the triangle here. To remind you, the law of cosines tells us the relationship between the sides of a triangle with side lengths a, b, and c:

$c^2=a^2+b^2-2ab\cos{C}$

Where C is the angle between sides a and b. c is typically the side we're trying to find, so on our triangle, we have

$c=AB\\a=BC=16\\b=CA=5\\C=m\angle C=61^{\circ}$

Substituting these values into the law of cosines:

$c^2=16^2+5^2-2(16)(5)\cos{61^{\circ}}\\c^2=256+25-160\cos{61^{\circ}}\\c^2=281-160\cos{61^{\circ}}\\c=\sqrt{281-160\cos{61^{\circ}}}\\c\approx 14.3$

6 0

3 years ago

Read 2 more answers

Which equation correctly applies the distributive property?

yuradex [85]

−3⋅(6.48)=(−3⋅6)+(−3⋅0.4)+(−3⋅0.08) correctly applies the distributive property

8 0

3 years ago

Read 2 more answers

Need help. Thank you :”)

Agata [3.3K]

It is the first choice. All you have to do is isolate A from 2A so you would divide by 2 and get A=bh/2

7 0

2 years ago

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Find the equation of all tangent lines having slope of -1 that are tangent to the curve y=(9)/(x+1)

fredd [130]

Answer:

$f(x)=\frac9{x+1}\\ f'(x)=-\frac9{(x+1)^2}\\ f'(x)=-1\ \iff\ -\frac9{(x+1)^2}=-1\ \to \ \frac9{(x+1)^2}=1\ \to \ (x+1)^2=9\\ |x+1|=3\ \to \ x+1=3\ \vee\ x+1=-3\\ x_1=2\ \vee\ x_2=-4\\ f(x_1)=f(2)=\frac9{2+1}=3\\ f(x_2)=f(-4)=\frac9{-4+1}=-3$

First tangent line:

$y=f'(x_1)\cdot (x-x_1)+f(x_1)\ \to \ y=-1(x-2)+3\ \to \ y=-x+5$

Second tangent line:

$y=f'(x_2)\cdot (x-x_2)+f(x_2)\ \to \ y=-1(x+4)-3\ \to \ y=-x-7$

Notice: slope of -1 means that both $f'(x_1), \ f'(x_2)$ are equal to -1, so $f'(x_1)=-1 \ and \ f'(x_2)=-1$

6 0

2 years ago

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