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3 years ago

If G is the midpoint of FH, find FG.

Mathematics

2 answers:

Mariana [72]3 years ago

6 0

Answer:

Step-by-step explanation:

15 is the answer because i looked it up

andreyandreev [35.5K]3 years ago

4 0

Answer:

FH/2

Step-by-step explanation:

FG = FH / 2

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NEED HELP ASAP PLEASE !!

Alexxandr [17]

Answer:

A, C, E

Step-by-step explanation:

The square root of 10 is approximately 3.16, which is greater than pi which is equal to about 3.14, which shows that C is correct as well. The square root of 11 is 3.31, and the square root of 5 added to the square root of 6 is equal to about 3.69, which is greater than the square root of 11.

6 0

3 years ago

two runners start 30kilometer apart and run toward each other. they meet after 3hours. One Runner runs 2 kilometers per hour fas

belka [17]

Let the slower runners speed be X kilometers per hour.

Then the faster runners speed would be X+2 kilometers per hour.

The formula for distance is Speed times time.

The distance is given as 30 kilometers and time is given as 3 hours.

Since there are two runners you need to add the both of them together.

The equation becomes 30 = 3x + 3(x+2)

Now solve for x:

30 = 3x + 3(x+2)

Simplify:

30 = 3x + 3x +6

30 = 6x + 6

Subtract 6 from each side:

24 = 6x

Divide both sides by 6:

x = 24/6

x = 4

The slower runner ran at 4 kilometers per hour.

The faster runner ran at 4+2 = 6 kilometers per hour.

3 0

3 years ago

Assume that y varies inversely

Marina86 [1]

Answer:

y = 2

Step-by-step explanation:

y varies inversely with x setup is:

y = k/x

7 = $\frac{k}{2/3}$ (find 'k')

k = 7/1 · 2/3

k = 14/3

use what you know about 'k' and 'x' to solve for 'y'

y = 14/3 ÷ 7/3 (remember to multiply by the reciprocal when dividing fractions)

y = 14/3 · 3/7

y = 2

3 0

3 years ago

Solve for j. -1j+1>2 Please help.

prohojiy [21]

You can use photomath for this

6 0

3 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 bin is constructed from 48 square feet of sheet metal, so:

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, the dimensions of the largest volume of such a bin is:

Length = 4 ft
Width = 4 ft
Height = 2 ft

And its volume is:

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

8 0

3 years ago