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2 years ago

What is the sum of the exterior angels of a 28 sided polygon ?

Mathematics

1 answer:

Inga [223]2 years ago

6 0

Answer: All polygons follow a rule that the sum of their exterior angles will equal 360 degrees.

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Solve for j.<br>-2j + 39 = 111<br>j=<br>

olga_2 [115]

Answer:

reeeeeerre

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

What is the surface area of the solid shown below?

Ugo [173]

Answer:

$\text{C. }156\:\mathrm{cm^3}$

Step-by-step explanation:

The surface area of the triangular prism consists of three rectangles and two triangles.

Rectangle One - 5 x 12 = 60 cm^2

Rectangle Two - 4 x 12 = 48 cm^2

Rectangle Three - 3 x 12 = 36 cm^2

Triangles - 2 x 1/2 x 3 x 4 = 12 cm^2

Thus, the total surface area of the solid is $60+48+36+12=\boxed{\text{C. }156\:\mathrm{cm^3}}$

8 0

3 years ago

Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule

Sergeeva-Olga [200]

Answer:

First term: 5

Fourth term: 5 1/2

Tenth term: 6 1/2

Step-by-step explanation:

Let's find the first, fourth and tenth terms of the arithmetic sequence described by the given rule:

A(n) = 5 + (n-1) (1/6)

First term:

A(1) = 5 + (1-1) (1/6)

A(1) = 5 + (0) (1/6)

A(1) = 5

Fourth term:

A(4) = 5 + (4-1) (1/6)

A(4) = 5 + (3) (1/6)

A(4) = 5 + 3/6 = 5 3/6 = 5 1/2 (simplifying)

Tenth term:

A(10) = 5 + (10-1) (1/6)

A(10) = 5 + (9) (1/6)

A (10) = 5 + 9/6 = 6 3/6 = 6 1/2 (simplifying)

5 0

3 years ago

Explain how the value of the 7 in 327,902 will change if you move it to the tens place

-BARSIC- [3]

The tens place will be where the zero is which the value would be 70 since the 7 is in the thousand place the value is 7,000

7 0

3 years ago

Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.

Romashka [77]

By definition, the average change of rate is given by:
$AVR = \frac{f(x2)-f(x1)}{x2-x1}$
We will calculate AVR for each of the functions.
We have then:

f(x) = x^2 + 3x interval: [-2, 3]:
$f(-2) = x^2 + 3x = (-2)^2 + 3(-2) = 4 - 6 = -2

f(3) = x^2 + 3x = (3)^2 + 3(3) = 9 + 9 = 18$
$AVR = \frac{-2-18}{-2-3}$
$AVR = \frac{-20}{-5}$
$AVR = 4$

f(x) = 3x - 8 interval: [4, 5]:
$f(4) = 3(4) - 8 = 12 - 8 = 4 f(5) = 3(5) - 8 = 15 - 8 = 7$
$AVR = \frac{7-4}{5-4}$
$AVR = \frac{3}{1}$
$AVR = 3$

f(x) = x^2 - 2x interval: [-3, 4]
$f(-3) = (-3)^2 - 2(-3) = 9 + 6 = 15

f(4) = (4)^2 - 2(4) = 16 - 8 = 8$
$AVR = \frac{8-15}{4+3}$
$AVR = \frac{-7}{7}$
$AVR = -1$

f(x) = x^2 - 5 interval: [-1, 1]
$f(-1) = (-1)^2 - 5 = 1 - 5 = -4

f(1) = (1)^2 - 5 = 1 - 5 = -4$
$AVR = \frac{-4+4}{1+1}$
$AVR = \frac{0}{2}$
$AVR = 0$

Answer:
these functions from the greatest to the least value based on the average rate of change are:
f(x) = x^2 + 3x
f(x) = 3x - 8
f(x) = x^2 - 5
f(x) = x^2 - 2x

5 0

3 years ago