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2 years ago

15

Simplify: 2 (13- | -11+2|) -|10 - 7|^2

1 answer:

jasenka [17]2 years ago

5 0

The answer will be -1.

I hope it helps. Let me know.

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The area of the circle is 144π m2<br><br> What is the diameter of the circle?

Over [174]

Answer:

Step-by-step explanation:

Remember that

Area of Circle = $\pi r^2$

$\pi = pi \\r = radius$

They told us that for this circle, the area is $144\pi m^2$

If you put it back into the formula for area of a circle we get:

Area of Circle = $\pi r^2$

$144\pi = \pi r^2$

Divide both sides by $\pi$ .

144 = $r^2$

Now we want to solve for r (r = radius).

You need to square root each side

$\sqrt{144} = \sqrt{r^2}$

$12 = r$

So now we know the radius, $r = 12$ metres

Also remember that

Diameter = $2r$

So,

Diameter = 2 times 12 = 24 metres

5 0

3 years ago

Determine whether y varies directly with x. If so, find the constant of variation k.

kumpel [21]

Direct variation is y=kx, inverse variation is y=k/x....

-5y=-2x divide both sides by -5

y=2.5x

So this is a direct variation with a constant of 2.5

5 0

3 years ago

What is 8.275 rounded to the nearest hundredth.

miskamm [114]

Answer:

8.3

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the slope of the line that passes through (5,12) and (9,7)

omeli [17]

The slope will be the difference between the y-coordinates divided by the difference between the x-coordinates:

then the slope of the line is:

$m=\frac{12-7}{5-9}=\frac{5}{-4}=-\frac{5}{4}$

8 0

1 year 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:

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

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

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0

2 years ago