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faltersainse [42]
3 years ago
9

What is the quotient (x3 + 6x2 + 11x + 6) ÷ (x2 + 4x + 3)? x + 2 x – 2 x + 10 x + 6

Mathematics
2 answers:
sveta [45]3 years ago
5 0
If you would like to solve (x^3 + 6x^2 + 11x + 6) / (x^2 + 4x + 3), you can do this using the following steps:

(x^3 + 6x^2 + 11x + 6) / (x^2 + 4x + 3<span>) = ((x + 1) * (x + 2) * (x + 3)) / ((x + 1) * (x + 3)) = x + 2
</span>
The correct result would be x + 2.
ratelena [41]3 years ago
4 0
Hello,

Answer A

x²+4x+3=(x+3)(x+1)

x^3+6x²+11x+6=(x+1)(x²+5x+6)

x²+5x+6=(x+2)(x+3)

so

x^3+6x²+11x+6=(x+1)(x+2)(x+3) (Horner's method)


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GEOMETRY PROBLEMS PLEASE HELP ME
SIZIF [17.4K]

Answer:

12) The measure of PQR is 36 since both angles MQP (25) and PQR should add up to MQR (61)

13) The graph shows that both angles are congruent so that means:

2x+8= x+22

2x= x+14

x=14

14) angle ADB and angle BDC since it shows that both angles sum up to 90 degrees (also known as complementary angles)

15) Angles ADE and angle FDE are supplementary because they both add up to 180 degrees (supplementary angles)

16) Angles ADC and FDE are vertical because they are opposite of each other

17) Angles ADE and FDE are also linear pairs because both of them are adjacent (they both have a common side as well as a common vertex)

Hope this helps! ;)

7 0
2 years ago
1 At the post office 55% of the packages ship as priority mail. If 160 packages are
ziro4ka [17]

Answer:

72

Step-by-step explanation:

45% did not get shiped so 160*.45=72

6 0
3 years ago
Is y=x^3 a solution of the differential equation yy'=x^5+y
shepuryov [24]

No; we have y=x^3\implies y'=3x^2. Substituting these into the DE gives

3x^5=x^5+x^3

which reduces to x^3=0, true only for x=0.

3 0
3 years ago
How do you solve this equation?
Travka [436]
Exponentail thingies

easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0

now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero

5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1


5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4

x=log₅1 and/or log₅4







second quesiton

same thing

1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero

2^x-8=9
2^x=8
x=3

2^x-2=0
2^x=2
x=1

x=3 or 1







first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3


7 0
3 years ago
Simplify the expression.<br> (-3+6)(-3+5)
Kipish [7]

Answer:

6

Step-by-step explanation:

first you do -3+6 which is 3.

then, u do -3+5 which is 2

multiply it and u get 6

5 0
3 years ago
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