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tatuchka [14]
3 years ago
5

Braden biked 10 8/9 miles on Monday. On Tuesday, he biked 3/ 4 as far as Monday. How many total miles did he bike over the two d

ays?
Mathematics
1 answer:
torisob [31]3 years ago
5 0

Answer:

19 1/18 miles.

Step-by-step explanation:

For Tuesday's ride we need to multiply 10 8/9 by 3/4.

10 8/9 = (9*10+8) / 9 =  98/9 miles.

98 / 9  * 3/4

= 98/3 * 1/4

49/6

= 8 1/6 miles.

Total miles over 2 days = 10 8/9 + 8 1/6

= 18 + 8/9 + 1/6

= 18 +  48/54 + 9/54

= 18 57/54

= 19 3/54

= 19 1/18 miles.

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Vladimir [108]
Isolate the variable by dividing each side by the factor that don’t contain the variable.
Answer: t < -1
4 0
3 years ago
A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder top. The company will make 1863 mailboxes t
forsale [732]

Answer:

3433 m²

Step-by-step explanation:

From the image, we have a rectangular box without cover and half a cylinder on top.

Formula for surface area of rectangular box with top is;

S = 2(lh + wh + lw)

From the image,

l = 0.6 m

w = 0.4 m

h = 0.55 m

Thus;

S = 2((0.6 × 0.55) + (0.4 × 0.55) + (0.6 × 0.4))

S = 1.58 m²

Now, since the top is not included for this figure, then;

Surface area of this rectangular box is;

S1 = 1.58 - (lw) = 1.58 - (0.4 × 0.6) = 1.34 m²

Surface area of a cylinder is;

S = 2πr² + 2πrh

r is radius and in this case = 0.4/2 = 0.2 m

h = 0.6

S = 2π(0.2² + (0.2 × 0.6))

S = 1.005 m²

Since it is half cylinder, then we have;

S2 = 1.005/2

S2 = 0.5025 m²

Total surface area; S_t = S1 + S2

S_t = 1.34 + 0.5025

S_t = 1.8425 m²

This is the surface area of one mail box.

Thus, for 1863 mailboxes, total surface area is;

S = 1863 × 1.8425 = 3432.5775 m²

Approximating to the nearest Sq.m gives;

S = 3433 m²

6 0
3 years ago
Bill is bisecting an angle with technology Which of the following describes his next step?
maw [93]

Answer:

  Mark point E where the circle intersects segment BC

Step-by-step explanation:

Apparently, Bill is using "technology" to perform the same steps that he would use with compass and straightedge. Those steps involve finding a point equidistant from the rays BD and BC. That is generally done by finding the intersection point(s) of circles centered at D and "E", where "E" is the intersection point of the circle B with segment BC.

Bill's next step is to mark point E, so he can use it as the center of one of the circles just described.

___

<em>Comment on Bill's "technology"</em>

In the technology I would use for this purpose, the next step would be "select the angle bisector tool."

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Step-by-step explanation:

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