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3 years ago

If f(x) = x^3-3, what is the average rate of change of f(x) over the interval [-4, 6]f?

Mathematics

2 answers:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer

42 would be the answer

Step-by-step explanation:

Vinil7 [7]3 years ago

6 0

42 is the answer your looking for.

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Find f(g(x)) and g(f(x)). f(x)=x²-2 and g(x)=√x+1

Archy [21]

Answer:

1. x - 1

2. $\sqrt{x^2-1}$

Step-by-step explanation:

$f(g(x)) = \sqrt{x+1}^2 - 2\\ = x + 1 - 2\\= x - 1$

$g(f(x)) =\sqrt{x^2-2+1}\\ = \sqrt{x^2-1}$

6 0

1 year ago

the perimeter of an isosceles trapezoid is 62 cm. if three sides are equal in length and the fourth side is 10 cm longer, find t

Alja [10]

Perimeter = sum of all the sides
let x = one side
longer side = x + 10
Perimeter = x + x + x + x + 10
62 = 4x + 10
4x = 52
x = 13
Longer side = 23

Area = (10+23)x2/10
= 6.6

5 0

2 years ago

Read 2 more answers

A gardener planted some seedlings at equal distances apart along a straight road. The 3rd seedling and the 5th seedling were 9/1

ratelena [41]

Answer:

Answer:

17 seedlings

Step-by-step explanation:

Given;

The seedlings are at an equal distance in a straight line.

Distance between 3rd seedling and 5th seedling = 9/10 m

So, distance between 3rd seedling and 4th seedling = 9/10 ÷ 2 = 9/20 m

So, the distance between two seedlings in a line = 9/20 m

The given distance between 2nd seedling and the last seedling = 6 3/4 m = 27/4 m

8 0

2 years ago

find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$

hope helpful :D

6 0

1 year ago

What’s measure of angle C?

Oduvanchick [21]

The angle is 60° because
B is a right angle and that adds to 90°, so dividing 90 by 3 would get 30

30 x 2 = 60

and to check the answer

x = 30

A=30
B=90
C=60

this adds to 180 which is correct because the angles in a triangle add to 180° :)

7 0

3 years ago