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Amiraneli [1.4K]
3 years ago
13

What is the complete factorization of the polynomial below?x^3 - 4x^2 + x-4​

Mathematics
1 answer:
postnew [5]3 years ago
5 0

Answer:

<h2>B. (x-4)(x+i)(x-i) </h2>

Step-by-step explanation:

x³-4x²+x-4=

=x²(x-4)+(x-4)

=(x-4)(x²+1)

=(x+4)(x-i²)

<h3>=(x-4)(x+i)(x-i) </h3>

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Andy reads 56 more pages of his book on Monday than Tuesday. He reads 125 pages on Monday. How many pages in all does he read on
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He read 181 pages total on Monday and Tuesday.
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in a class, 60% of the students are girls. there are 19 girls in the class. how many students are there? how many boys are there
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60/40 =19/? 19 ×40÷60=13 there is 13 boys
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3 years ago
Simplify the expression<br> 65-4^2 (10-2)+37 <br> please show all work
BartSMP [9]

Hello there!

65-4^2 (10-2)+37 \\

Explanation:

↓↓↓↓↓↓↓↓↓↓↓↓↓

I gave you a hint.

You had to used PEMDAS.

P-Parenthesis

E-Exponents

M-Multiply

D-Divide

A-Add

S-Subtract

First you had to calculate with parenthesis first.

65-4^2(10-2)+37

65-4^2*8+37

Then you calculate with exponents.

4^2=4*4=16

65-16*8+37

You had to multiply and divide you had go to left to right.

16*8=128

65-128+37

Then you add or subtract you had go to left to right.

65-128+37=-26

=-26

Answer⇒⇒⇒⇒⇒⇒=-26

Hope this helps!

Thank you for posting your question at here on Brainly.

Have a great day!

-Charlie

6 0
3 years ago
Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

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What num com between 4 and 6?
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Answer:

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Step-by-step explanation:

1,2,3,4,5,6

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