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svetoff [14.1K]
3 years ago
8

Which statement(s) could be used to prove RTS TRU?

Mathematics
2 answers:
Ostrovityanka [42]3 years ago
6 0

Answer:

SAS

Step-by-step explanation:

Law Incorporation [45]3 years ago
4 0

all of those r correct :)

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An event manager recorded the number of people in different age groups who attended a music concert:
Vesnalui [34]

Answer:

Age Group Number of People 18–24 80 25–31 120 32–38 40 39–45 20

The data in each bar is how many people in that group.

So 18-24 has 80 people

25-31 has 120 people

32 -38 has 40 people

39-45 has 20 people

Step-by-step explanation:

3 0
2 years ago
Help me with this<br>Please Help​
Arte-miy333 [17]
Answer:
80mm

Step by step explanation:

To convert centimeters to millimeters, all you have to do is multiply the number of centimeters by 10.
3 0
1 year ago
Read 2 more answers
Which side lengths form a right triangle?
Natalija [7]

Answer:

this is kinda confusing..So ill just answer the first question.

<h3>Explanation: The Pythagorean Theorem gives us a2 + b2 = c2 for a right triangle, where c is the hypotenuse and a and b are the smaller sides.</h3>

hope it helps.

5 0
3 years ago
-4w+18w+2=10-6w <br><br> What is w?
kiruha [24]

Answer:

w = .5

Step-by-step explanation:

-4w + 18w+2=10 -6w

14w + 2 -2 = 10 -2 -6w

14w + 2w = 8 - 2w + 2w

16w = 8

16w/16 = 8/16

w = .5

8 0
3 years ago
Read 2 more answers
An adult brain is about 140 mm wide and divided into two sections (called "hemispheres" although the brain is not truly spherica
Sedbober [7]

Answer: 3.61×10^5 A

Step-by-step explanation: Since the brain has been modeled as a current carrying loop, we use the formulae for the magnetic field on a current carrying loop to get the current on the hemisphere of the brain.

The formulae is given below as

B = u×Ia²/2(x²+a²)^3/2

Where B = strength of magnetic field on the axis of a circular loop = 4.15T

u = permeability of free space = 1.256×10^-6 mkg/s²A²

I = current on loop =?

a = radius of loop.

Radius of loop is gotten as shown... Radius = diameter /2, but diameter = 65mm hence radius = 32.5mm = 32.5×10^-3 m = 3.25×10^-2m

x = distance of the sensor away from center of loop = 2.10 cm = 0.021m

By substituting the parameters into the formulae, we have that

4.15 = 1.256×10^-6 × I × (3.25×10^-2)²/2{(0.021²) + (3.25×10^-2)²}^3/2

4.15 = 13.2665 × 10^-10 × I/ 2( 0.00149725)^3/2

4.15 = 1.32665 ×10^-9 × I / 2( 0.000058)

4.15 × 2( 0.000058) = 1.32665 ×10^-9 × I

I = 4.15 × 2( 0.000058)/ 1.32665 ×10^-9

I = 4.80×10^-4 / 1.32665 ×10^-9

I = 3.61×10^5 A

3 0
3 years ago
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