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Vitek1552 [10]
3 years ago
8

Joelle is working on a school report in a word processor. She wants to place an oversized image which is 30.51 \text{ cm}30.51 c

m30, point, 51, start text, space, c, m, end text wide so that it appears centered on the page. The width of the page is 21.59 \text{ cm}21.59 cm21, point, 59, start text, space, c, m, end text. In the program, horizontal positions to the right of the leftmost edge of the page are positive. What horizontal position for the left edge of the image should Joelle use to center the image?
Mathematics
1 answer:
ehidna [41]3 years ago
3 0

Answer:

the horizontal position for the left edge of the image is -4.46 cm.

Step-by-step explanation:

The horizontal midpoint of the page is:

21.59 / 2 = 10.79 cm

The horizontal midpoint of the image is given as:

30.51 / 2 = 15.25 cm

These midpoints must be on the same point in the axis.

By taking the leftmost edge of the paper to be point zero on the axis, then, the distance accommodated by the paper is 10.795 cm.  

The distance that goes beyond this leftmost edge is computed as;

This is on the negative side on the axis.

Thus the horizontal position for the left edge of the image is -4.46 cm.

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An athlete knows that when she jogs along her neighborhood​ greenway, she can complete the route in 10 minutes. It takes 20 minu
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Answer:

10 miles per hour.

Step-by-step explanation:

Let x represent athlete's walking speed.

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\text{Distance}=\text{Rate}\cdot \text{Time}

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While walking, we will get D_{\text{walking}}=x\frac{\text{miles}}{\text{hour}}\cdot \frac{1}{3}\text{hour}

D_{\text{walking}}=\frac{x}{3}

While jogging, we will get D_{\text{jogging}}=(x+5)\frac{\text{miles}}{\text{hour}}\cdot \frac{1}{6}\text{hour}

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Since athlete is covering same distance while walking and jogging, so we can equate both expressions as:

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