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2 years ago

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An article fills 1/2 of a magazine page. A corresponding photo takes up 3/8 of the article. How much of the page is taken up by

the photo

2 answers:

erastova [34]2 years ago

6 0

Answer:

$\frac{3}{16}$ part of a page.

Step-by-step explanation:

An article fills $\frac{1}{2}$ part of a magazine page.

Hence, if the total magazine page is x then $\frac{x}{2}$ of the magazine page is covered by an article.

Now, given that, a corresponding photo takes up $\frac{3}{8}$ part of the article.

Therefore, $\frac{x}{2}\times \frac{3}{8} =\frac{3x}{16}$ i.e. $\frac{3}{16}$ part of the total page is taken up by the photo. ( Answer )

tankabanditka [31]2 years ago

6 0

<u>Answer:</u>

An article fills 1/2 of a magazine page. The photo takes 3/16th of the page

<u>Solution:</u>

A corresponding photo takes up 3/8 of the article

We have to find out how much of the page is taken by photo

This can be calculated by finding out the space taken by photo in the total article space

Space taken by article = 1/2 page

Space taken by photo= 3/8th of article

3/8th of article means 3/8th of 1/2

$=\frac{3}{8} \times \frac{1}{2}=\frac{3}{16}$

Hence the corresponding photo takes 3/16 of the magazine page

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Artemon [7]

<h2>The missing measures of the angles are:</h2>

$m\angle 1= 60^{\circ}\\\\m\angle 2= 39^{\circ}\\\\m\angle 3= 21^{\circ}\\\\m\angle 4 = 39^{\circ}\\\\m\angle 5 = 21^{\circ}$

<u>Given</u>:

$m\angle Z = 138^{\circ}$

<em><u>Note the following:</u></em>

An equilateral triangle has all its three angles equal to each other. Each angle = $60^{\circ}$ .

This implies that, since $\triangle WXY$ is equilateral, therefore, $m\angle Y = m\angle XWY = m\angle XYW = 60^{\circ}$

Base angles of an isosceles triangle are congruent to each other.

This implies that, since $\triangle WZY$ is isosceles, therefore, $m\angle 3 = \angle 5$

Applying the above stated, let's find the measure of each angle:

Find $m\angle 1$

$m\angle 1 = 60^{\circ}$ (an angle in an equilateral triangle equals 60 degrees)

Find $m\angle 2$

$m\angle 2 = 60 - m \angle 3$

$m\angle 2 = 60 -\frac{1}{2}(180 - 138)$ $(Note: \frac{1}{2}(180 - 138) = 1 $ base $ angle $ of $ \triangle WZY)$

$m\angle 2 = 60 -21\\\\m\angle 2 = 39^{\circ}$

Find $m\angle 3$ and $m\angle 5$ (base angles of isosceles triangle WZY)

$m\angle 3 = \frac{1}{2}(180 - 138) (1 $ base $ angle $ of $ \triangle WZY)$

$m\angle 3 = \frac{1}{2}(42) \\\\m\angle3 = 21^{\circ}$

$m\angle3 = m\angle 5$ (base angles of isosceles triangle are congruent)

Therefore,

$m\angle5 = 21^{\circ}$

Find $m\angle 4$

$m\angle 4 = 60 - m\angle 5$

Substitute

$m\angle 4 = 60 - 21\\\\m\angle 4 = 39^{\circ}$

The missing measures of the angles are:

$m\angle 1= 60^{\circ}\\\\ m\angle 2= 39^{\circ}\\\\m\angle 3= 21^{\circ}\\\\m\angle 4 = 39^{\circ}\\\\m\angle 5 = 21^{\circ}$

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