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2 years ago

Benito wrote the equations shown about the figure. Explain Benito’s errors.

Mathematics

1 answer:

Minchanka [31]2 years ago

7 0

Answer:

Benito's error was using the equal sign (=) instead of the congruency symbol (≅).

Explanation:

Benito's error was using the equal sign (=) instead of the congruency symbol (≅).

The congruency symbol (≅) means that the elements (segments, angles or figures in general) have the same measure, i.e. they have equal lengths for the segments or equal measure for the angles.

For instance, it is an error saying that the segment AB is equal to the segment BC because, as you clearly see in the picture, they are not same; they have the same length but they are joining different points, that makes them different in essence, although they have the same length. They would be equal only if they are the same figure.

In mathematics, you must not say that two different segments or two different angles are equal but they are congruent, which means that their lengths are equal. The use of equal is reserved for numbers and variables, not for figures like segment, points, angles, polygons.

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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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Miranda bought a square frame that has an area of 30 square inches what is the approximate side lenght of the frame

Alexxandr [17]

Area of a square equals side squared

(A = s²)

30 = s²

√30 = s

You can use a calculator to find out that √30 ≈ 5.48

or you can use estimation (√25 < √30 < √36) which means that √30 is between 5 and 6.

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2 years ago

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Jill is standing at the base of her apartment building. She measures the angle of elevation to the top of a nearby tower to be 4

almond37 [142]

1) From the measure of 40°, you can write:

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⇒x=100/tan(40°) = 119,18 m

2) From the measure of 30°, you can write

tan(30°) = y / 119,18, where y is the height from the roof of Jill's building to the top of the tower.

Then, y = tan(30°) * 119,18 = 68,81 m

3) The height of Jill's building is 100 - 68,81 = 31,19 m

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2 years ago

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Lera brought a watermelon that weighed b kg to a party. Nick brought a box of candies. How much is the weight of the box of cand

Dennis_Churaev [7]

Answer:

1.5b (kg)

Step-by-step explanation:

Let's begin by listing out the variables we were given:

weight of the watermelon = b (kg),

weight of watermelon = (2/5) * weight of candies

weight of candies = 1 ÷ (2/5) = 1 ÷ 0.4

weight of candies = 2.5b (kg)

How much is the weight of the box of candies greater than the weight of the watermelon is given by:

weight of the box of candies - weight of watermelon= 2.5b - b = <u>1.5b</u> (kg)

<u>Therefore, the weight of the box of candies is greater than the weight of the watermelon by 1.5b (kg) </u>

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2 years ago

On a number line, the coordinates of X, Y, Z, and W are , , , and , respectively. Find the lengths of the two segments be

Arturiano [62]

Complete Question:

On a number line, the coordinates of X, Y, Z, and W are −8, −5, 4, and 6, respectively. Find the lengths of the two segments below. Then tell whether they are congruent. $\overline{XY}$ and $\overline{ZW}$