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schepotkina [342]
2 years ago
12

Click on all measurements that are equal to 0.001km wide

Mathematics
1 answer:
Morgarella [4.7K]2 years ago
6 0

Answer:

100cm, 1m are equal to 0.001km

Step-by-step explanation:

0.001 is 1 thousandths and a kilometer is 1,000 meters so it would be 1m/1000m, 100 centimeter because one meter is one hundred centimeters.

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How many solutions for linear equation 3x + 5y = 25 can be found so that −20 < y < 10, and x is an integer?
ch4aika [34]

Answer:

50

Step-by-step explanation:

3x = 25 - 5y

-100 < 5y < 50

-50 < -5y < 100

-25 < 25 - 5y < 125

-8.333 < x < 41.667

41+8+1 = 50

5 0
3 years ago
Compute the norm of v when v = (-8,7,4)​
dexar [7]

Answer:

the norm of V = =11.357816691601..

Step-by-step explanation:

\left\vert \overrightarrow{V} \right\vert  =\sqrt{\left( -8\right)^{2}  +\left( 7\right)^{2}  +\left( 4\right)^{2}  }

      =\sqrt{64+49+16}

      =\sqrt{129}

      =11.357816691601...

4 0
1 year ago
Ruby works in a bike shop where she makes repairs. She needs to order tires for bikes and quads, which have different kinds of t
valkas [14]

Each bike has two tires, while each quad has four tires. Therefore the number of tires Ruby has to other is given by the number of tires used by the bikes added by the number of tires used by the quads. This is shown on the expression below:

\text{tires}=2\cdot b+4\cdot q

5 0
9 months ago
Solve 11b^2-9=-68 using the square root method
Vladimir79 [104]

Answer:

b=i*\sqrt{\frac{59}{11} } or -i*\sqrt{\frac{59}{11} }

Step-by-step explanation:

11b^2-9=-68

11b^2=-59

b^2=-59/11

b=i*\sqrt{\frac{59}{11} } or -i*\sqrt{\frac{59}{11} }

"i" in this case is an imaginary number, equal to \sqrt{-1\\}

if you haven't learned about these yet, something is wrong with the question

3 0
3 years ago
two mountain bikers leave from the same parking lot and head in opposite directions on two different trails. the first rider goe
sveta [45]

Applying the required <em>rule </em>or <em>theorem</em>, it can be concluded that the second biker is <u>farther</u> from the <em>parking lot</em>. The distance of the bikers to the <em>parking lot</em> are:

i. First biker = 17.0 km

ii. Second biker = 20.22 km

The <u>path</u> of travel of both bikers would form a triangle. Applying the <u>Pythagoras</u> theorem to the path of the <em>first</em> biker would give his <u>distance</u> from the starting point. While applying the <u>cosine</u> rule to the path of <em>second</em> rider would gives his <u>distance</u> to the starting point.

Thus,

a. <u>To determine the distance of the first biker from the parking lot.</u>

Let the required <em>distance </em>be represented by x. Applying the Pythagoras theorem, we have:

hyp^{2} = adj 1^{2} + adj 2^{2}

x^{2} = 8^{2} + 15^{2}

   = 64 + 225

   = 289

x = \sqrt{289}

  = 17

x = 17 km

Thus, the <u>first</u> biker is 17.0 km from the <em>starting</em> point.

b. <u>To determine the distance of the second biker from the parking lot.</u>

Let the required <em>distance</em> be represented by x. So that applying the cosine rule, we have:

c^{2} = a^{2} + b^{2} - 2ab Cos θ

x^{2} = 8^{2} + 15^{2} - 2(15*8) Cos (180 - 20)

    = 64 + 225 - 240 Cos 160

    = 289 - 240 * -0.5

x^{2} = 289 + 120

   = 409

x = \sqrt{409}

 = 20.2237

x = 20.22 km

Thus, the <u>second</u> biker is 20.22 km from the <em>starting</em> point.

Therefore, the second biker is <u>farther</u> from the <em>parking lot</em>.

A sketch of the path of travel for the two bikers is attached for more clarifications.

Visit: brainly.com/question/22699651

7 0
2 years ago
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