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Elenna [48]
3 years ago
13

A scientist uses a submarine to study ocean life.

Mathematics
2 answers:
Alexus [3.1K]3 years ago
6 0

Answer:

31.3 feet

Step-by-step explanation:

-23.6+13.4 = -10.2

-10.2 - 21.1 = -31.3

Setler [38]3 years ago
5 0

Answer:

31.3 feet

Step-by-step explanation:

0 -23.6 + 13.4 - 21.1 = -31.3

So to get back to sea level she must ascend 31.3 feet.

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What is the answer ?
sergiy2304 [10]

Answer:

First box: 2

Second box: 30

Third box: 42

Step-by-step explanation:

5 sixes means 5 · 6, which equals 30. An unknown amount of sixes equals 12, which is 2 because 2 · 6 = 12.

30 + 12 = 42

7 · 6 = 42, so 42 is the answer.

I hope this helped! :)

4 0
3 years ago
Read 2 more answers
4. Find the time duration for the following in your notebook: (a) 1:20 a.m. to 8.05 a.m. (b) 3:50 p.m. to 6:25 p.m. (C) 11.40 a.
padilas [110]

(a) 6 hours and 45 minutes

(b) 2 hours and 35 minutes

(c) 2 hours and 40 minutes

(d) 3 hours and 20 minutes

4 0
2 years ago
HELPPPP
deff fn [24]
I think that the best answer is A
4 0
3 years ago
Plz help me with this ASAP
VLD [36.1K]

Answer:

Step-by-step explanation:

plug all in

1) 0 when x=0

2) 1*6-1²=6-1=5

3) 6*2-2²=12-4=8

4)6*3-3²=18-9=9

5) 6*4-4²=24-16=8

6)6*5-5²=30-25=5

7) 6*6-6²=36-36=0

6 0
3 years ago
Match each set of vertices with the type of triangle they form.
Andrew [12]

Answer:  The calculations are done below.


Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

4 0
3 years ago
Read 2 more answers
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