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2 years ago

Will a 7/32 inch diameter pin fit into a reamed hole with a diameter of 0.215 inches?

Mathematics

1 answer:

lana [24]2 years ago

6 0

Answer:

no ... the pin will not fit the reamed hole

Step-by-step explanation:

The decimal equivalent of 7/32 is 0.21875. This is larger than the reamed

hole: 0.215

and so the answer is no ... the pin will not fit the reamed hole.

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Which of the following equations could describe a square pyramid? Select all that apply.

Katena32 [7]

C. w= 3V/lh

D. V/B=h/3

E. V=w^2h/3

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2 years ago

Makayla is taking a math course and is working with the perimeter of rectangles. She knows the perimeter and length of her recta

jeka94

Answer:

$W=\dfrac{P}{2}-L$

Step-by-step explanation:

$P=2L+2W \qquad\text{given}\\\\P-2L=2W \qquad\text{subtract $2L$}\\\\\dfrac{P-2L}{2}=W \qquad\text{divide by 2}\\\\W=\dfrac{P}{2}-L \qquad\text{carry out the division}$

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2 years 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 $2$ .

Apparently $\displaystyle c = \frac{16}{10} = \frac{8}{5}$ works. $c = 1.6$ while $\displaystyle \frac{1}{c} = \frac{5}{8} = 0.625$ .

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2 years ago

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