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Sphinxa [80]
3 years ago
9

Use the information to answer questions 8 and 9.

Mathematics
1 answer:
inessss [21]3 years ago
8 0

Answer:

18

Step-by-step explanation:

the factors and prime factorization of 36 and 54. The biggest common factor number is the GCF number. So the greatest common factor 36 and 54 is 18.

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nasty-shy [4]
38,000 is the answer
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6, 12, 18 and 24 are the first four multiples of 6.<br><br> What are the next 2 multiples?
kondor19780726 [428]

Answer:

30 and 36

Step-by-step explanation:

six times two equals 12

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and six times four equals 24.

to get the next two multiples, You need to multiply 6 by 5, and then 6

6 * 5 = 30 and 6 * 6 = 36

3 0
3 years ago
A cylinder has a total volume of 678.59 and a height of 6m. what is the diameter of the cylinder? please help and show work
RoseWind [281]
Volume =
\pi {r}^{2} h
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7 0
3 years ago
Solve x^2+20x+100=50
uysha [10]
1. You con solve the quadratic equation x^2+20x+100=50<span> by following the proccedure below:
 
 2. Pass the number 50 from the right member to the left member. Then you obtain:
 
 x^2+20x+100-50=0
</span><span> x^2+20x+50=0
</span><span> 
 3. Then, you must apply the quadratic equation, which is:
 
 x=(-b±√(b^2-4ac))/2a
 
 </span><span>x^2+20x+50=0
</span><span> 
 a=1
 b=20
 c=50
 

  4. Therefore, when you substitute the values into the quadratic equation and simplify ir, you obtain that the result is:
 
  -10</span>±5√2   (It is the last option).
6 0
3 years ago
Consider the following. f(x) = x5 − x3 + 6, −1 ≤ x ≤ 1 (a) Use a graph to find the absolute maximum and minimum values of the fu
kykrilka [37]

ANSWER

See below

EXPLANATION

Part a)

The given function is

f(x) =  {x}^{5}  -  {x}^{3}  + 6

From the graph, we can observe that, the absolute maximum occurs at (-0.7746,6.1859) and the absolute minimum occurs at (0.7746,5.8141).

b) Using calculus, we find the first derivative of the given function.

f'(x) = 5 {x}^{4} - 3 {x}^{2}

At turning point f'(x)=0.

5 {x}^{4} - 3 {x}^{2}  = 0

This implies that,

{x}^{2} (5 {x}^{2}  - 3) = 0

{x}^{2}  = 0 \: or \: 5 {x}^{2}  - 3 = 0

x =  - \frac{ \sqrt{15} }{5}   \: or \: x = 0 \:  \: or \: x =\frac{ \sqrt{15} }{5}

We plug this values into the original function to obtain the y-values of the turning points

(   -  \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( 6 \sqrt{15}  +750)) \:and \:  (0, - 6) \: and\: (   \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( - 6 \sqrt{15}  +750))

We now use the second derivative test to determine the absolute maximum minimum on the interval [-1,1]

f''(x) = 20 {x}^{3}  - 6x

f''( -  \frac{ \sqrt{15} }{5} ) \:   <  \: 0

Hence

(   -  \frac{ \sqrt{15} }{5}  , \frac{1}{125} ( 6 \sqrt{15}  + 750))

is a maximum point.

f''( \frac{ \sqrt{15} }{5} ) \:    >  \: 0

Hence

(     \frac{ \sqrt{15} }{5}  , \frac{1}{125} (- 6 \sqrt{15}  + 750))

is a minimum point.

f''(0) \: =\: 0

Hence (0,-6) is a point of inflexion

4 0
3 years ago
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