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Nostrana [21]
3 years ago
13

What is 16500 rounded to the nearest ten thousand?

Mathematics
1 answer:
frosja888 [35]3 years ago
4 0
16500 Rounded to the nearest ten thousand is 20,000
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someone please help, write in point-slope form an equation of the line that passes through the point (8, 9) with slope 7.
Bess [88]

Answer:

y=7x-47

Step-by-step explanation:

(8, 9) = (x_1,y_1) \\ m = 7 \\ y  - y_1 = m(x - x _1) \\ y - 9 = 7(x - 8)

y - 9 = 7x - 56 \\ y = 7x - 56 + 9 \\ y = 7x - 47

7 0
3 years ago
Last Question...Please help if you can...(25 points if you answer this)
tresset_1 [31]

Answer:

  • Annie- 17 hours, Brian- 3 hours

Step-by-step explanation:

Let the number of hours Annie travelled be a, and same of Brian be b

<u>Then we have</u>

  • a + b = 20
  • a = 5b + 2

<u>Substituting a in the first equation</u>

  • 5b + 2 + b = 20
  • 6b = 20 - 2
  • 6b = 18
  • b = 3

<u>Then finding a</u>

  • a = 20 - 3 = 17

<u>So the answer is:</u>

  • Annie- 17 hours, Brian- 3 hours
5 0
4 years ago
Olivia has an average score 90 on her five French tests if she earns a score of 98 on the sixth test what will he average score
SVETLANKA909090 [29]

Answer:

91.33

Step-by-step explanation:

90x5=450, add 98 to get 548. Since there are 6 tests, divide 548 by 6 and you get 91.33

7 0
4 years ago
Help me to answer this question pl​s
attashe74 [19]

Problem 1

Draw a straight line and plot P anywhere on it. Use the compass to trace out a faint circle of radius 8 cm with center P. This circle crosses the previous line at point Q.

Repeat these steps to set up another circle centered at Q and keep the radius the same. The two circles cross at two locations. Let's mark one of those locations point X. From here, we could connect points X, P, Q to form an equilateral triangle. However, we only want the 60 degree angle from it.

With P as the center, draw another circle with radius 7.5 cm. This circle will cross the ray PX at location R.

Refer to the diagram below.

=====================================================

Problem 2

I'm not sure why your teacher wants you to use a compass and straightedge to construct an 80 degree angle. Such a task is not possible. The proof is lengthy but look up the term "constructible angles" and you'll find that only angles of the form 3n are possible to make with compass/straight edge.

In other words, you can only do multiples of 3. Unfortunately 80 is not a multiple of 3. I used GeoGebra to create the image below, as well as problem 1.

8 0
2 years ago
Graph the following piecewise function and then find the range.
Lelechka [254]
C, [1,109). Hope this helps you m8
4 0
4 years ago
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