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3 years ago

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There is a mound of g pounds of gravel in a quarry. Throughout the day, 400 pounds of gravel are added to the mound. Two orders

of 900 pounds are sold and the gravel is removed from the mound. At the end of the day, the mound has 1,500 pounds of gravel. Write the equation that best describes the situation.

1 answer:

Anni [7]3 years ago

3 0

Answer:

The equation that best describes the situation is:

$g +400 -1800 = 1,500$

Step-by-step explanation:

The initial amount of gravel is g.

$g$

Then we know that 400 pounds are added

$g +400$

Two orders of 900 pounds are sold and the gravel is removed from the mound. This is:

$g +400 -2 * 900$

$g +400 -1800$

At the end of the day, the mound has 1,500 pounds of serious. This is:

$g +400 -1800 = 1,500$

The equation that best describes the situation is:

$g +400 -1800 = 1,500$

And

$g= 1500 +1800 - 400\\\\g=2900$

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3 years ago

A 40-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and the chain

daser333 [38]

Answer:

a) W₁ = 78400 [J]

b)Wt = 82320 [J]

Step-by-step explanation:

a) W = ∫ f*dl general expression for work

If we have a chain with density of 10 Kg/m, distributed weight would be

9.8 m/s² * 10 kg = mg

Total length of th chain is 40 m, and the function of y at any time is

f(y) = (40 - y ) mg where ( 40 - y ) is te length of chain to be winded

At the beggining we have to wind 40 meters y = 0 at the end of the proccess y = 40 and there is nothing to wind then:

f(y) = mg* (40 - y )

W₁ = ∫f(y) * dy ⇒ W₁ = ∫₀⁴⁰ mg* (40 - y ) dy ⇒ W₁ = mg [ ∫₀⁴⁰ 40dy - ∫₀⁴⁰ ydy

W₁ = mg [ 40*y |₀⁴⁰ - 1/2 * y² |₀⁴⁰ ⇒ W₁ = mg* [ 40*40 - 1/2 (40)² ]

W₁ = mg * [1/2] W₁ = 10*9,8* ( 800 )

W₁ = 78400 [J]

b) Now we can calculate work to do if we have a 25 block and the chain is weightless

W₂ = ∫ mg* dy ⇒ W₂ = ∫₀⁴⁰ mg*dy ⇒ W₂ = mg y |₀⁴⁰

W₂ = mg* 40 = 10*9.8* 40

W₂ = 3920 [J]

Total work

Wt = W₁ + W₂ ⇒ Wt = 78400 + 3920

Wt = 82320 [J]

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3 years ago

(1 point) A fish tank initially contains 15 liters of pure water. Brine of constant, but unknown, concentration of salt is flowi

Drupady [299]

Answer:

a. $\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

b. $x(t) = 6\frac{2}{3} \cdot c \cdot t$

c. $c = \dfrac{3}{8} \ g/L$

Step-by-step explanation:

a. The volume of water initially in the fish tank = 15 liters

The volume of brine added per minute = 5 liters per minute

The rate at which the mixture is drained = 5 liters per minute

The amount of salt in the fish tank after t minutes = x

Where the volume of water with x grams of salt = 15 liters

dx = (5·c - 5·c/3)×dt = 20/3·c = $6\frac{2}{3} \cdot c \cdot dt$

$\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

b. The amount of salt, x after t minutes is given by the relation

$\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

$dx = 6\frac{2}{3} \cdot c \cdot dt$

$x(t) = \int\limits \, dx = \int\limits \left ( 6\frac{2}{3} \cdot c \right) \cdot dt$

$x(t) = 6\frac{2}{3} \cdot c \cdot t$

c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;

$x(10) = 25 \ grams(15 \ in \ liters) = 6\frac{2}{3} \times c \times 10$

$6\frac{2}{3} \times c =\dfrac{25 \ grams }{10}$

$c =\dfrac{25 \ g/L }{10 \times 6\frac{2}{3} } = \dfrac{25 \ g/L}{10 \times \dfrac{20}{3} } =\dfrac{3}{200} \times 25 \ g/L= \dfrac{75}{200} \ g/L = \dfrac{3}{8} \ g/L$

$c = \dfrac{3}{8} \ g/L$

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2 years ago

What is the equation of a quadratic function P with rational coefficients that

Phoenix [80]

Answer:

$P(x) = x^2-6x+58$

Step-by-step explanation:

Quadratic function is function of the form $P(x)=ax^2+bx+c$ where a, b and c are real numbers and $a\neq 0$

A number is said to be a rational number if it can be written in the form $\frac{u}{v}$ where u and v are integers and $v\neq 0$ .

A number is said to be a complex number if it can be written in the form u+iv where u and v are real numbers.

A number 'a' is said to be a zero of a function P(x) if $P(a)=0$

We know that if zero of a fraction is of form u+iv then its other zero is u-iv.

As 3+7i is a zero of the function, 3-7i is also its zero.

So, given equation is $\left ( x-(3-7i) \right )\left ( x-(3+7i) \right )$

$P(x)=\left ( x-(3-7i) \right )\left ( x-(3+7i) \right )\\=x^2-x(3+7i)-x(3-7i)+(3-7i)(3+7i)\\=x^2-3x-7ix-3x+7ix+9+49\\=x^2-6x+58$

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3 years ago