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3 years ago

What is the perimeter of the figure shown below? a. 20 ft c. 12 ft b. 6 ft d. 24 ft

Mathematics

2 answers:

Contact [7]3 years ago

4 0

Answer:

P = 24 ft

Step-by-step explanation:

Step 1: FInd how many sides there are.

There are 12 sides to this figure.

Step 2: Find how many feet ONE SIDE equals.

ONE SIDE = 2 ft

Step 3: Add 2 ft 12 times(multiply 2 by 12)

2+2+2+2+2+2+2+2+2+2+2+2 = 12 OR

2 x 12 = 24

<em>Therefore, the perimeter of this figure is 24 ft.</em>

Hope this helped!! :)

Ilya [14]3 years ago

3 0

Answer:

d) 24

Step-by-step explanation:

there are 12 sides on the star so multiply the length of the known side times 12.

2 x 12 = 24

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1. m+ 10< 16

sleet_krkn [62]

Answer:

$1.\ m$

Step-by-step explanation:

1. Subtract 10 from both sides. Then:

$m+ 10< 16\\ m+ 10-10< 16-10\\m$

2. Divide both sides by -2. Notice that the direction of the symbol of the inequality will change. Then:

$-2g\geq8\\\\\frac{-2g}{-2}\geq\frac{8}{-2}\\\\g\leq-4$

3. Add 22 to both sides. Then:

$y-22< 19\\ y-22+22$

4. Divide both sides by -7. Notice that the direction of the symbol of the inequality will change. Then:

$-7b>-28\\\\\frac{-7b}{-7}>\frac{-28}{-7}\\\\b$

5. Subtract 30 from both sides. Then:

$h+30$

6. Divide both sides by -5. Notice that the direction of the symbol of the inequality will change. Then:

$-5x$

7. Subtract 13 from both sides. Then:

$t+13>22\\ t+13-13>22-13\\t>9$

8. Add 12 to both sides. Then:

$w-12< 16\\ w-12+12$

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3 years ago

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3 years ago

Read 2 more answers

Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.