What is the lateral area of this regular octagonal pyramid?

the class kept track of rainy and sunny days. During the 54 days, classified rainy or sunny, the ratio of rainy days is 7 to 14.

Rom4ik [11]

There are 36 sunny days

6 0

1 year ago

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Dev has 9 shells. Zoe has 55 shells, Zoe gives some shells to Dev. Now zoe has 3 times as many shells as dev. How many shells do

Goryan [66]

Answer:

number of shells zoe gives to dev = 7

number of shells zoe is left with = 55-7= 48

number of shells dev has = 9+7=16

Step-by-step explanation:

let the initial number of shells that dev has be A, and initial number of shells that zoe has be B.

let the number of shells that zoe gives to dev be x.

after giiving x shells zoe is left with 3 times the number of shell as that of dev.

therefore number of shells with zoe = 3×number of shells with dev.

number of shells with zoe = initial shells - x = 55-x

number of shells with dev = initial shells + number of shells he gets

= 9+x

therefore (55-x)=3×(9+x)

55-x = 27+3x

55-27=3x+x

4x = 28

x= 7

therefore number of shells zoe gives to dev = 7

number of shells zoe is left with = 55-7= 48

number of shells dev has = 9+7=16

6 0

1 year ago

I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

makkiz [27]

Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .