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2 years ago

Find the value of x. (7x - 1)º (6x - 1)º

Mathematics

1 answer:

dangina [55]2 years ago

6 0

Answer:

Step-by-step explanation:

7x - 1 + 6x - 1 = 180

13x - 2 = 180

13x = 182

x = 14

7(14) - 1 = 98 - 1= 97

6(14) - 1 = 84 - 1 = 83

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Is 4/67 grater that 5/777

Anni [7]

Yes because if you divide 4 by 67 you get about 0.0597 and if you divide 5 by 777 then you get about 0.0064 and 0.0597 is greater than 0.0064

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3 years ago

Please Answerrr! I need help ASAPP!

MArishka [77]

Step-by-step explanation:

5×6 = 30 and 30×8 = 240 divide that by three you get 80

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2 years ago

What is the correlation

RideAnS [48]

Answer:

type=positive

Correlation refers to the degree of correspondence or relationship between two variables.

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2 years ago

HELP ME, PLZ. 100 POINTS !!!

marta [7]

Answer:

x=2√5/5

x=-2√5/5

Step-by-step explanation:

13√(2√5/5)^2-(2√5/5)^4+9√(2√5/5)^2+(2√5/5)^4=16

13√(4×5/25)-(16×25/625)+9√(4×5/25)+(16×25/625)

13√(4/5)-(16/25)+9√(4/5)+(16/25)

13√(4/25)+9√(36/25)

13×2/5+9×6/5

26/5+54/5

16=16 (proven)

3 0

2 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

2 years ago