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kow [346]
3 years ago
6

The length of a school hallway is 115 meters how many kilometers long is the hallway

Mathematics
1 answer:
Levart [38]3 years ago
4 0

Answer:

0.115 kilometers

Step-by-step explanation:

One kilometer is equal to 1,000 meters, so 115 meters is

115/1,000 kilometers,  or 0.115 kilometers

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9 * 9 ^ 2 x + 3 equal 27 * 3 ^ x - 2​
Amiraneli [1.4K]

3^2*3^2 (2x+3)=3^2+4x+6=3^4x+8

3^3*3^x-2=3^x+1

4x+8=x+1

3x=-7

x=-7/3

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3 years ago
What’s it called when 3^2 ? how do you say that
Aleks [24]

"Three to the power of two"

or

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or

"Three Squared"

8 0
4 years ago
Read 2 more answers
What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?
Akimi4 [234]

Answer:

25.6 units

Step-by-step explanation: From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).

First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:

d=\sqrt{(x_2-x_1)^{2} +(y_2-y_1)^{2}}

where

(x_1,y_1) are the coordinates of the first point

(x_2,y_2) are the coordinates of the second point

- For AB:

d=\sqrt{[1-(-5)]^{2}+(4-4)^2}

d=\sqrt{(1+5)^{2}+(0)^2}

d=\sqrt{(6)^{2}}

d=6

- For BC:

d=\sqrt{(3-1)^{2} +(-4-4)^{2}}

d=\sqrt{(2)^{2} +(-8)^{2}}

d=\sqrt{4+64}

d=\sqrt{68}

d=8.24

- For AC:

d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}

d=\sqrt{(3+5)^{2} +(-8)^{2}}

d=\sqrt{(8)^{2} +64}

d=\sqrt{64+64}

d=\sqrt{128}

d=11.31

Next, now that we have our lengths, we can add them to find the perimeter of our triangle:

p=AB+BC+AC

p=6+8.24+11.31

p=25.55

p=25.6

We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.

Read more on Brainly.com - brainly.com/question/12560433#readmore

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3 years ago
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Svetradugi [14.3K]

Answer:

Lol

Step-by-step explanation:

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3 years ago
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Help me please!!!!!!!!!!!!
goldenfox [79]

Answer: I think it's no, sorry if I can't help

Step-by-step explanation:

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