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3 years ago

Can some one help me out please

Mathematics

1 answer:

liberstina [14]3 years ago

3 0

Answer:

If t is the number of years after 2003, then t = 0 in 2003.

Step-by-step explanation:

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4(0.5f−0.25)=6+f what is f I am so lost when we were doing this in class

muminat

4 (0.5f - 0.25) = 6 + f
2.0f - 1.0 = 6.0 + f
1.00f = 6.0 + f
f = -5.0
hope this is right,
~ Harley Quinn~

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3 years ago

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Identify if the proportion is true or false. 4 to 11 = 12 to 33.

yuradex [85]

Answer:

True

Step-by-step explanation:

Take 4/11 and you get 0.363636363636, which is the same if you take 12/33. So the proportion of the two is the same.

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Which quadrant point is in (2,5)

n200080 [17]

Answer:

Sweets are sold loose, or pre-packed in 120g bags.

The 120 g bags are £1.49 each.

The loose sweets are £0.89 for 100g.

By calculating the price per gram, determine which is better value. Show your working.

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2 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

Hence, the amount of oil was decreasing at 69300 barrels, yearly

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2 years ago

Simplify 3x + 6x^2 - 5x - x^2

just olya [345]

Answer: -2x + 5x^2

Step-by-step explanation:

3x + 6x^2 - 5x - x^2 = x(3-5) + x^2(6-1) = -2x + 5x^2

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3 years ago

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