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tatiyna
3 years ago
13

A turtle is 6 feet below the surface of the sea. If his position can be recorded as −6 feet, what would the position of 0 repres

ent?
6 feet in the sea
6 feet above the water
On the surface of the sea
At the bottom of the sea
help me asap
Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
5 0

Answer:

The answer is On the surface of the sea. I just got it right. Please choose me the Brainliest. Thanks Guys for your support

Step-by-step explanation:

Anarel [89]3 years ago
3 0
On the surface of the sea. My best guess because the surface is neither above or below
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Help please :) about to post more too
yKpoI14uk [10]

Answer:

Lets say test tubes = t, and beakers = b

1 pack of (t) is $4 less than 1 pack of (b)

Since i have no prior information we are going to use variables for this equation:

1t (1 pack of test tubes) is $4 less than 1b (1 set of beakers)

so to quantify the equation, we have 8t and 12b.

if b is a number that IS quantifiable such as $5 we can easily figure out this answer.

Lets use and example that 1 set of beakers is $8, if we multiply $8 by 12 (the number of sets of beakers), we get: 96

Using the same example, if 1t is $4 less than 1b than 1t = $4. So, if we multiply $4 by 8 (the amount of packs of test tubes), we get: 32

If you take both of those numbers: 96, and 32 and you divide them you get 3. so that means that 1t = 3b

Answer = 1t = 3b

This may not be correct due to the little information that i got however i hope that, that works out for you :)

5 0
3 years ago
Solve y=x^2-5 for x​
Natasha_Volkova [10]

Answer:

x^2 - 5 = y

x^2= y + 5

x \:  =  \:  \sqrt{y + 5}

7 0
3 years ago
Need a correct answer please
zloy xaker [14]

Answer       x=14         y=1

Step-by-step explanation:

3 0
2 years ago
If you place a 89-foot ladder against the top of a 80-foot building, how many feet will
Nuetrik [128]

Answer:

\large \boxed{\text{39 ft}}

Step-by-step explanation:

Let x be the distance from the foot of the ladder to the base of the building.

We have a right triangle, so we can use Pythagoras' Theorem.

\begin{array}{rcl}80^{2} + x^{2}& = & 89^{2} \\6400+ x^{2} & = & 7921\\x^{2} & = & 1521\\x & = & \sqrt{1521}\\& = &\mathbf{39}\\\end{array}\\\text{The foot of the ladder will be $\large \boxed{\textbf{39 ft}}$ from the base of the building,}

8 0
3 years ago
Find the distance CD rounded to the nearest tenth C=(5,6) D=(2,2)
Dmitriy789 [7]
2,2 is 3,4 away from 5,6 and 3²+4²=25=5² so CD=5
3 0
2 years ago
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