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3 years ago

Find the x- and y-intercepts of the equation. 3x − 3y = 12

Mathematics

1 answer:

Romashka [77]3 years ago

6 0

X=-4-y there you go enjoy

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Is the following statement true or false?<br> A plane has an endpoint.

rjkz [21]

The answer is false because a plane goes on infinitively

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2 years ago

Find the rate of markup if a book cost the retailer 17.50 and sells for $24.95. Round the the nearest tenth if needed

CaHeK987 [17]

Mark up = profit ÷ cost

Let M = mark up

Let p = profit

Let c = cost

M = p/c

Take it from here.

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3 years ago

20 - 4x = 2(4 - 8x) This needs solving

Olenka [21]

20 - 4x = 2(4 - 8x)

Make sure to do the parentheses first or factor out first

20 - 4x = 8 - 16x

add 16x to both sides

20 + 12x = 8

Subtract 20

12x = -12

Divide by 8

x = -1

8 0

3 years ago

Read 2 more answers

Jana finished 1/4 of her homework at school and 2/3 of it before dinner. How much of her homework has she done

Galina-37 [17]

Answer:

the answer is 11/12

Step-by-step explanation:

well she has done 11/12 of her work so she just needs to do i more question baccically

5 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