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Free_Kalibri [48]

3 years ago

6

(a)Tom bought 12 pounds of sugar for $6 . How many dollars did he pay per pound of sugar? (b)It takes 55 pounds of seed to compl

etely plant a 6 -acre field. How many pounds of seed are needed per acre? If necessary, round your answers to the nearest hundredth.

1 answer:

Gala2k [10]3 years ago

3 0

(a) 0.50 per pound, you do 6 divided by 12
(b) 9.17 pounds per field, you do 55 divided by 6

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My estimate was 2,808 is that a reasonable estimate and why

vova2212 [387]

<span>My estimate was 2,808 is that a reasonable estimate and why

312*9

9*2=18>>9*10=90>>9*300=2700

2700+18+90=>>2,808</span>

5 0

4 years ago

Use signed numbers to solve. A helicopter descended 1200 feet, rose 800 feet, and then descended 450 feet. What was the net gain

DENIUS [597]

B. Lost 850 ft
800-450= 350
1200-350=850

I hope I helped :D

6 0

3 years ago

A pyramid has a square base with sides of length s the height of The pyramid is equal to 1/2 Of the length of a side on the base

DENIUS [597]

Answer:

The correct answer is s³/6

Step-by-step explanation:

<u>Points to remember</u>

Volume of pyramid = (a²h)/3

Where a - side of base

h - height of pyramid

It is given that, a pyramid has a square base with sides of length s the height of The pyramid is equal to 1/2 Of the length of a side on the base

<u>To find the formula:-</u>

Here side length a = s and

height h = 1/2(s) = s/2

Volume = (a²h)/3

= (s² * s/2)/3

= s³/6

The correct answer is s³/6

3 0

3 years ago

Read 2 more answers

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six

kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

$V = Length \times Width \times Height \\\\$

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

$Width = 3 - 2x \\\\$

The length of the shaded region is given by

$Length = \frac{1}{2} (5 - 3x) \\\\$

So, the volume of the box becomes,

$V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\$

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

$\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\$

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

$a = 18 \\\\b = -38 \\\\c = 15 \\\\$

$x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\$

Volume of the box at x= 1.59:

$V = \frac{1}{2} (5 – 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\$

Volume of the box at x= 0.53:

$V = \frac{1}{2} (5 – 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3$

The volume of the box is maximized when x = 0.53 cm

Therefore,

x = 0.53 cm

Maximum volume = 1.75 cm³

7 0

3 years ago

If i=√-1 , what is i^2 equal to ?

Alja [10]

Answer:

-1

Step-by-step explanation:

If i = √-1, then i^2 = -1.

5 0

3 years ago