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NeX [460]
3 years ago
7

Find a. Round it to the nearest tenth

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
6 0

Answer:

42637646e6r7r7e7464664

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I really need help with this Triangle problem. ​
dedylja [7]

Answer:

12

Step-by-step explanation:

Here we use SOHCAHTOA

Seeing as we have the hypotenuse and we are working to find out the opposite, we use SOH

Sin(x)= Opposite/Hypotenuse

Sin(45) = x/12\sqrt{2} \\

×12 root 2

sin(45) times 12 root 2=  12

x=12

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3 years ago
How to find the accurate estimate of the volume?
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3 years ago
Type your answer and then click or tap Done.<br> Simplify -(x-2y) - y.
AlladinOne [14]

Answer:

Step-by-step explanation:

x(x-2y)-(y-x)2  

Final result :

 -y2

Step by step solution :

Step  1  :

Equation at the end of step  1  :

 x • (x - 2y) -  (y - x)2

Step  2  :

2.1     Evaluate :  (y-x)2   =    y2-2xy+x2  

Final result :

 -y2

4 0
3 years ago
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What is the number below that subtracted 7 from four times it multiplied and equals 1?
Dvinal [7]

Answer:

a.4

Step-by-step explanation:

because

4 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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