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Lemur [1.5K]
4 years ago
13

If p(x) = x 3 + x 2+ root 5 x + root 5 , then find p( – 5 )

Mathematics
1 answer:
Morgarella [4.7K]4 years ago
6 0

Answer:

-100-4√5

Step-by-step explanation:

Given the polynomial expression

p(x) = x³+x²+√5x + √5

p(-5) = (-5)³+(-5)² + √5(-5) + √5

p(-5) = -125 + 25 -5√5 + √5

p(-5) = -100-4√5

This gives the required answer

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The matrix A= (−3 0 1, 2 −4 2, −3 −2 1) has one real eigenvalue. Find this eigenvalue, its multiplicity, and the dimension of th
Aliun [14]

Answer:

a) -4

b) 1

c) 1

Step-by-step explanation:

a) The matrix A is given by:

A=\left[\begin{array}{ccc}-3&0&1\\2&-4&2\\-3&-2&1\end{array}\right]

to find the eigenvalues of the matrix you use the following:

det(A-\lambda I)=0

where lambda are the eigenvalues and I is the identity matrix. By replacing you obtain:

A-\lambda I=\left[\begin{array}{ccc}-3-\lambda&0&1\\2&-4-\lambda&2\\-3&-2&1-\lambda\end{array}\right]

and by taking the determinant:

[(-3-\lambda)(-4-\lambda)(1-\lambda)+(0)(2)(-3)+(2)(-2)(1)]-[(1)(-4-\lambda)(-3)+(0)(2)(1-\lambda)+(2)(-2)(-3-\lambda)]=0\\\\-\lambda^3-6\lambda^2-12\lambda-16=0

and the roots of this polynomial is:

\lambda_1=-4\\\\\lambda_2=-1+i\sqrt{3}\\\\\lambda_3=-1-i\sqrt{3}

hence, the real eigenvalue of the matrix A is -4.

b) The multiplicity of the eigenvalue is 1.

c) The dimension of the eigenspace is 1 (because the multiplicity determines the dimension of the eigenspace)

3 0
3 years ago
The diameter of a circle is 38 centimeters. What is the circle's area?
adoni [48]

Answer:

area of a circle ⭕= πr²

radius = diameter ÷2;

radius = 38÷2

radius = 19

3.14 * 19²

= 1133.54 cm²

5 0
3 years ago
WILL MARK BRAINLIST!!!
MAVERICK [17]
D. 200%

if 8 is a whole (100%) and 16 is twice of that, you can conclude that her speed increased by 200%
6 0
3 years ago
Rationalize the denominator<br><br> 11/Square root of 5 - square root of 6
olasank [31]

Answer:

-11\sqrt{5}-11\sqrt{6}

Step-by-step explanation:

The given radical expression is;

\frac{11}{\sqrt{5}-\sqrt{6}}

We multiply both the numerator and the denominator by the conjugate of

\sqrt{5}-\sqrt{6} which is \sqrt{5}+\sqrt{6}

\frac{11(\sqrt{5}+\sqrt{6})}{(\sqrt{5}-\sqrt{6})(\sqrt{5}+\sqrt{6})}

The denominator is now difference of two squares;

\frac{11(\sqrt{5}+\sqrt{6})}{(\sqrt{5})^2-(\sqrt{6})^2}

Simplify:

\frac{11(\sqrt{5}+\sqrt{6})}{5-6}

\frac{11(\sqrt{5}+\sqrt{6})}{-1}

-11(\sqrt{5}+\sqrt{6})

-11\sqrt{5}-11\sqrt{6}

8 0
4 years ago
What is the length of JK¯¯¯¯¯, to the nearest tenth of a millimeter? 4.5 mm 5.6 mm 8.3 mm 19.9 mm A horizontally-aligned scalene
Leokris [45]

Answer:

5 = side a

3 = side c

angle B = 62 degrees

JK is side b

Use the Law of Cosines to find the length of side b.

b^2 = a^2 + c^2 − 2ac cos(B)

b^2 = (5^2 + 3^2) − ( 2 x 5 x 3) cos(62°)

b^2 = (25 + 9) − 30 x 0.469

b^2 = 34 - 30 x 0.469

b^2 = 34 - 14.084

b^2 = 19.916

b = √19.916

b = 4.463

Rounded to the nearest tenth is 4.5 mm.

Hope this helps!

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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