Find the domain of goh. g(x) = -x h(x) = 2x²+2x+3

yarga [219]

925 rounded to the nearest hundred is 900 becuase first we identify the hundreds digit which in this case is 9. Second,we identify the next smallest place value (the digit to the right of the hundreds place) which in this case is 2. Is that digit greater than or equal to five? No - we round Down. The hundreds digit is stays the same but every digit after it becomes a zero.

4 0

3 years ago

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In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank

fredd [130]

Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) $Amount \ of \, water \ remaining \ in \, the \ tank \ is \ \frac{x^3(6-\pi) }{6}$

Step-by-step explanation:

1) Here we have;

First tank A

Volume of tank = x³

The volume of the sphere = $\frac{4}{3} \pi r^3$

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = $\frac{4}{3} \pi \frac{x^3}{8}$ = $\frac{1}{6} \pi x^3$

For tank B

Volume of tank = x³

The volume of the spheres = $8 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 2·D = x therefore;

r = x/4 and the volume of the sphere is thus;

volume of the spheres = $8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}$

For tank C

Volume of tank = x³

The volume of the spheres = $64 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 4·D = x therefore;

r = x/8 and the volume of the sphere is thus;

volume of the spheres = $64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}$

Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The volume of the spheres = $512 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = $512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}$

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = $x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}$

$Amount \ of \ water \, remaining \, in \, the \ tank = \frac{x^3(6-\pi) }{6}$ .

5 0

3 years ago

Solve<br> 3(2x - 5) = 9(10 - x)

alisha [4.7K]

Answer:

x=7

Step-by-step explanation:

3(2x - 5) = 9(10 - x)

6x-15=90-9x

6x+9x=90+15

15x=105

x=7

4 0

3 years ago

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I need in depth help answering

Volgvan

What is the question

8 0

2 years ago

1/4+2/3-1/2= how do I solve this

Ann [662]

Answer:

5/12

Step-by-step explanation:

First you add 1/4+2/3.

You need to change the denominator so that they are equal to each other

If you multiply the 1/4 by 3/3 you will get 3/12 which is equal to 1/4

Then you need to do the same to 2/3. This time you need to multiply it by 4/4 to get the same denominator which will be 2/3*4/4=8/12

Then you add 3/12+8/12=11/12. You don't need to add the denominator.

After this you will need to subtract 11/12 and 1/2. This time you need to only change the denominator for 1/2.

You multiply the denominator and numerator by 6/6 to 1/2 and you will get 6/12 then you are going to subtract 11/12-6/12 and you will get 5/12

6 0

3 years ago