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EastWind [94]
4 years ago
8

A one-hour mathematics class is divided up into the schedule shown in the table below.

Mathematics
2 answers:
Doss [256]4 years ago
7 0

Answer:

the correct answer would be C: 18 minutes

Step-by-step explanation:

Nat2105 [25]4 years ago
4 0

Answer:

It would be 8 minutes maybe but I don't have enough information to say that is the answer.

Step-by-step explanation:

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Use the law of cosines to find the length of a
borishaifa [10]

Answer:

Step-by-step explanation:

You don't need the Law of Cosines, the Law of sines if what you need. You can't use the Law of Cosines because in order to find side a, you would need the length of side c and you don't have it. Using the Law of Sines is appropriate, knowing that angle B = 55:

\frac{sin55}{175}=\frac{sin42}{a} and solving for a:

a=\frac{175sin42}{sin55} so

a = 143.0

4 0
3 years ago
The length of a chord is equal to its distance to the center of the circle. A second chord in the same circle is twice as long a
Rama09 [41]

Answer:

The distance from the center of the circle to the longer chord is twice smaller than the distance from the center to the shorter chord.

Step-by-step explanation:

The length of the chord AB is the same as the distance OC from the center to the cord. Let OC=2x, then CA=x. By the Pythagorean theorem, the radius r of the circle is

r^2=OC^2+AC^2,\\ \\r^2=(2x)^2+x^2=5x^2,\\ \\r=\sqrt{5}x.

The length of the arc ED is 4x.

Consider right triangle EFO. In this triangle, EF=2x, EO=r, then the distance OF is

OF^2=OE^2-EF^2,\\ \\OF^2=5x^2-(2x)^2=x^2,\\ \\OF=x.

The distance from the center of the circle to the longer chord is twice smaller than the distance from the center to the shorter chord.

5 0
4 years ago
Find first 4 terms and the 10th term of each question. 1)n2+3 2) 2n2
irina [24]

Answer:

1) 4, 7, 12 and 19

Tenth term = 103

2) 2, 8, 18, 32

Tenth term =200

Step-by-step explanation:

Given the nth term of the sequence

1) f(n)= n²+3

When n = 1

f(1) = 1²+3

f(1) = 4

When n = 2

f(2) = 2²+3

f(2) =4+3

f(2) = 7

When n = 3

f(3) = 3²+3

f(3) = 9+3

f(3)=12

When n =4

f(4) = 4²+3

f(4) = 16+3

f(4) = 19

The first four terms are 4, 7, 12 and 19

f(10)= 10²+3

f(10) = 103

Tenth term is 103

2) f(n) = 2n²

When n = 1

f(1) = 2(1)²

f(1) = 2

When n = 2

f(2) = 2(2)²

f(2) = 2(4)

f(2) = 8

When n= 3

f(3)= 2(3)²

f(3) = 2(9)

f(3) = 18

When n= 4

f(4) = 2(4)²

f(4) = 2(16)

f(4) =32

The first four terms are 2, 8, 18, 32

f(10) = 2(10)²

f(10) =2(100)

f(20) = 200

The tenth term is 200

5 0
3 years ago
21+[36/ (12-6)+2-5]x3
vampirchik [111]
The answer is 30 i believe i rewrote it 3 times to check
3 0
4 years ago
How does a digit in the ten thousand place compare to a digit in the thousands place
dezoksy [38]
It is ten times bigger than the other
4 0
3 years ago
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