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3 years ago

9

Write an equation in slope-intercept form for the line that passes through the point (-8, 2) and is perpendicular to the line y

= 4/5x + 10.

1 answer:

tamaranim1 [39]3 years ago

3 0

Answer:

y = -5x/4 - 8

Step-by-step explanation:

Two lines that are perpendicular have slopes that are opposite reciprocals.

Since the slope of the given line is 4/5, the opposite reciprocal of that is -5/4. So the equation has a slope of -5/4.

Since we know the coordinate of a point of the other equation, we can plug that into the point-slope form equation y - y1 = m(x - x1):

y - 2 = -5/4(x-(-8))

Simplify:

y - 2 = -5/4(x+8)

y - 2 = -5x/4 - 10

So the equation in slope-intercept form is:

y = -5x/4 - 8

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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

Pls help:Find all the missing elements:

Yuliya22 [10]

Answer:

<h3>B = 48.7° , C = 61.3° , b = 12</h3>

Step-by-step explanation:

In order to find B we must first angle C

To find angle C we use the sine rule

That's

$\frac{ |a| }{ \sin(A) } = \frac{ |c| }{ \sin(C) }$

From the question

a = 15

A = 70°

c = 14

So we have

$\frac{15}{ \sin(70) } = \frac{14}{ \sin(C) }$

$\sin(C) = \frac{14 \sin(7 0 ) }{15}$

$C = \sin^{ - 1} ( \frac{14 \sin(70) }{15} )$

C = 61.288

<h3>C = 61.3° to the nearest tenth</h3>

Since we've found C we can use it to find B.

Angles in a triangle add up to 180°

To find B add A and C and subtract it from 180°

That's

A + B + C = 180

B = 180 - A - C

B = 180 - 70 - 61.3

<h3>B = 48.7° to the nearest tenth</h3>

To find b we can use the sine rule

That's

$\frac{ |a| }{ \sin(A) } = \frac{ |a| }{ \sin(B) }$

$\frac{15}{ \sin(70) } = \frac{ |b| }{ \sin(48.7) }$

$|b| = \frac{15 \sin(48.7) }{ \sin(70) }$

b = 11.9921

<h3>b = 12.0 to the nearest tenth</h3>

Hope this helps you

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3 years ago

You invest $5000 into an account that has a 2.7% annual interest rate and is compounded quarterly. Approximately how long will i

allsm [11]

Since the interest is compounded, we will have to use the compound interest formula.

We Weill plug 7500 in for A, because that's the amount of money that we want to have at the end of some amount of time.

5000 will go in for P because that's the starting amount.

2.7% will be converted into a decimal percentage form. You can do this by dividing by 100, which you will get .027, and then plug that in for r, the rate.

Since the interest is compounded quarterly, n = 4.

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Answer:

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