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3 years ago

11

Find the area of a regular octagon with apothem K and side of 10.

1 answer:

Svet_ta [14]3 years ago

4 0

Answer:

area = (n * side length * apothem) / 2

where "n" is number of sides

area = 8 * 10 * K / 2

area = 80K / 2

https://www.1728.org/polygon.htm

Step-by-step explanation:

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Area of a lake is 9/20 a square mile. The width is 1/5 the length. What is the length and the width.

katovenus [111]

Say, length =x mileWidth =. x/5 milex*x/5=9/2020x*x=45x*x=45/20 =9/4x=3/2 mileLength =3/2 milesWidth=. 3/10 miles

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3 years ago

What is the value for y?<br> Enter your answer in the box.

lapo4ka [179]

Answer:

y = 14

Step-by-step explanation:

The sum of the angles in a triangle add to 180 degrees

50 + 2x+ 5y + 10 = 180

We know that 2x = 50 since they are the base angles of sides that have the same length, they must be the same

50 + 50 + 5y+10 =180

Combine like terms

110 + 5y =180

Subtract 110 from each side

110+5y-110 = 180-110

5y = 70

Divide each side by 5

5y/5 =70/5

y = 14

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3 years ago

Read 2 more answers

5. Decide whether each equation is true for all, one, or no values of x.

S_A_V [24]

C doesn’t have a value , x equals 0 , x=0

6 0

3 years ago

Write a solution in Interval Notation - (you don't have to help me on all, 1 or 2 is fine c: )

Gemiola [76]

QUESTION 1

The given inequality is

$|m|-2>0$

We group like terms to get,

$|m|>2$

This implies that,

$-m>2$ or $m>2$ .

We simplify the inequality to get,

$m$ or $m>2$ .

We can write this interval notation to get,

$(-\infty,-2)\cup (2,+\infty)$ .

QUESTION 2

$|x-4|-3\:>\:5$ .

We group like terms to get,

$|x-4|\:>\:5+3$ .

$|x-4|\:>\:8$

We split the absolute value sign to get,

$-(x-4)\:>\:8$ or $x-4\:>\:8$

This implies that,

$x-4\:$ or $x-4\:>\:8$

$x\:$ or $x\:>\:8+4$

$x\:$ or $x\:>\:12$

We can write this interval notation to get,

$(-\infty,-4)\cup (12,+\infty)$ .

QUESTION 3

The given inequality is

$|6+9x|\leq 24$

We split the absolute value sign to obtain,

$-(6+9x)\leq 24$ or $(6+9x)\leq 24$

This simplifies to

$6+9x\ge -24$ and $6+9x\leq 24$

$9x\ge -24-6$ and $9x\leq 24-6$

$9x\ge -30$ and $9x\leq 18$

$x\ge -\frac{10}{3}$ and $x\leq 2$

$-\frac{10}{3}\leq x\leq2$

We write this in interval form to get,

$[-\frac{10}{3},2]$

QUESTION 4

The given inequality is

$|1-5a|>29$

We split the absolute value sign to get,

$-(1-5a)>29$ or $1-5a>29$

This simplifies to,

$1-5a\:$ or $1-5a\:>\:29$

This implies that,

$-5a\:$ or $-5a\:>\:29-1$

$-5a\:$ or $-5a\:>\:28$

$a\:>\:6$ or $a\:$

We write this in interval notation to get,

$(-\infty,-\frac{28}{5})\cup (6,+\infty)$

7 0

3 years ago

A rectangular field will be fenced on all four sides. There will also be a line of fence across the field, parallel to the short

Dimas [21]

Answer:

150m by 300 m

Step-by-step explanation:

150×300=45000

all sides add up to 900m

150+150+300+300

3 0

4 years ago