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2 years ago

7

The surface of the Dead Sea is 1312 feet below sea level. use an integer to express the surface of the Dead Sea. Hurry please I

need this before 7:40 pacific tomorrow

1 answer:

aleksklad [387]2 years ago

8 0

The answer would be -1312

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C= 5/9 (F−32)

Grace [21]

Answer:

C = 5/9 (F - 32)

This is a temperature scale change comparison.

By this we understand, that ,

1°C = 5/9°F

Which is equal to, 5/9°F = 1°C

These two statements are satisfied by I. and II.

So, the correct answers are, I. and II.

If my answer helped, kindly mark me as the brainliest!

Thank You!!

5 0

2 years ago

√37 < 5 1/4 (true or false)

Alex787 [66]

OK so on the top problem, the first thing you want to do is find the √37, once you know the answer to that you can compare it to 5 1/4 then it will be much easier to see is in fact √37 less than 5 1/4 or not.

On the bottom you want to know if 15 over the square root of ten is greater than 8.38 so what you want to do is to first find the square root of ten. after you know this then you put that number under 15. then you can compare what ever 15÷√10 is to 8.83. if you need any more info let me know.

4 0

3 years ago

Round 53.73 to the nearest whole number. A) 53 B) 53.7 C) 54 D) 54.7

liubo4ka [24]

53.73 rounded to the nearest whole number would be C) 54.

***hope this helps:) brainliest? ;)

3 0

2 years ago

The ratio of an object's weight on Earth to its weight on Mars is 5 to 2. How much would a man, who weighs 145 pounds on Earth,

butalik [34]

To answer this youd have to set up a proportion. 5/2 is the ratio of weight on earth and mars respectively. The other side has to have the same ratio where 125lbs is the weight on earth and x is the weight on mars. The equation would look like 5/2=125/x. Youd have to cross multiply, and would get (125)(2)=5x. Solve for x to find the weight on mars

6 0

2 years ago

A bin is constructed from sheet metal with a square base and 4 equal rectangular sides. if the bin is constructed from 48 square

kondaur [170]

This is a problem of maxima and minima using derivative.

In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:

V = length x width x height

That is:

$V = xxy = x^{2}y$

We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:

Surface area of the square base = $x^{2}$

Surface area of the rectangular sides = $4xy$

Therefore, the total area of the cube is:

$A = 48 ft^{2} = x^{2} + 4xy$

Isolating the variable y in terms of x:

$y = \frac{48- x^{2} }{4x}$

Substituting this value in V:

$V = x^{2}( \frac{48- x^{2} }{x}) = 48x- x^{3}$

Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:

$\frac{dv}{dx} = 48-3x^{2} =0$

Solving for x:

$x = \sqrt{\frac{48}{3}} = \sqrt{16} = 4$

Solving for y:

$y = \frac{48- 4^{2} }{(4)(4)} = 2$

Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

$V = (4^{2} )(2) = 32 ft^{3}$

8 0

2 years ago