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Ann [662]
3 years ago
9

Which of the following is one of the factors of the polynomial

Mathematics
1 answer:
RUDIKE [14]3 years ago
7 0

Answer:

5x-2

Step-by-step explanation:

15x^2-x-2

15x^2+5x-6x-2

factorize

5x(3x+1)-2(3x+1)

Sol: (3x+1)*(5x-2)

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David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below. GFC of 80, 32, and 48: 16 GCF o
BlackZzzverrR [31]

Answer: The correct statements are

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

David applied the distributive property.

Step-by-step explanation:

GCF = Greatest common factor

1) GCF of coefficients : (80,32,48)

80 = 2 × 2 × 2 × 2 × 5

32 = 2 × 2 × 2 × 2 × 2

48 = 2 × 2 × 2 × 2 × 3

GCF of coefficients : (80,32,48) is 16.

2) GCF of variables :(b^4,b^2,b^4)

b^4= b × b × b × b

b^2 = b × b

b^4 =b × b × b × b

GCF of variables :(b^4,b^2,b^4) is b^2

3) GCF of c^3 and c: c is not the GCF of the polynomial. The variable c is not common to all terms, so a power of c should not have been factored out.

4) 80b^4-32b^2c^3+48b^4c

=16b^2(5b^2-2c^3+3b^2c)

David applied the distributive property.

7 0
3 years ago
Read 3 more answers
Allied Corporation is trying to sell its new machines to Ajax. Allied claims that the machine will pay for itself since the time
kvasek [131]

Answer:

z(s) is in the acceptance region. We accept H₀  we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

Step-by-step explanation:

We must evaluate the differences of the means of the two machines, to do so, we will assume a CI  of 95%, and as the interest is to find out if the new machine has better performance ( machine has a bigger efficiency or the new machine produces more units per unit of time than the old one) the test will be a one tail-test (to the left).

New machine

Sample mean                  x₁ =    25

Sample variance               s₁  = 27

Sample size                       n₁  = 45

Old machine

Sample mean                    x₂ =  23  

Sample variance               s₂  = 7,56

Sample size                       n₂  = 36

Test Hypothesis:

Null hypothesis                         H₀             x₂  -  x₁  = d = 0

Alternative hypothesis             Hₐ            x₂  -  x₁  <  0

CI = 90 %  ⇒  α =  10 %     α = 0,1      z(c) = - 1,28

To calculate z(s)

z(s)  =  ( x₂  -  x₁ ) / √s₁² / n₁  +  s₂² / n₂

s₁  = 27     ⇒    s₁²  =  729

n₁  = 45    ⇒   s₁² / n₁    = 16,2

s₂  = 7,56   ⇒    s₂²  = 57,15

n₂  = 36     ⇒    s₂² / n₂  =  1,5876

√s₁² / n₁  +  s₂² / n₂  =  √ 16,2  + 1.5876    = 4,2175

z(s) = (23 - 25 )/4,2175

z(s)  =  - 0,4742

Comparing z(s) and  z(c)

|z(s)| < | z(c)|  

z(s) is in the acceptance region. We accept H₀  we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

The very hight dispersion of values s₁ = 27 is evidence of frecuent values quite far from the mean

3 0
3 years ago
the sum of three times a number and 2 less than 4 times the same number is 15. which of the following equations could be used to
Ulleksa [173]
To solve for a number x,

3x+4x-2=15
7x-2=15
7x=17
x=2 3/7
or x=~2.43

was one of your answer choices either of the first two equations?
4 0
3 years ago
Solve for x. -3.5x = -35
r-ruslan [8.4K]

Answer:

- 3.5x =  - 35 \\ x =  \frac{ - 35}{ - 3.5}(  -  \div  -  =  + ) =  \frac{35}{35 \times .1}  \\  =  \frac{1}{.1}  = 10 \\ x = 10 \\ thank \: you

7 0
3 years ago
Read 2 more answers
What mathematical process is the following formula performing: C2*D3?
tiny-mole [99]
Answer is D because C^2 ×D^2 the * mean ×
8 0
3 years ago
Read 2 more answers
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