Solve the system of equations 3x+2y=14 2x-4y=4

Answers

Answer 1

Answer: 3x + 2y = 14....multiply by 2
2x - 4y = 4
----------------
6x + 4y = 28 (result of multiplying by 2)
2x - 4y = 4
----------------add
8x = 32
x = 32/8
x = 4

2x - 4y = 4
2(4) - 4y = 4
8 - 4y = 4
-4y = 4 - 8
-4y = -4
y = -4/-4
y = 1

solution is (4,1)

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Help needed please.........

Answers

this is slope!!is it true???????

A house cost $120,000 when it was purchased. The value of the house increases by 10% each year. Find the rate of growth each month.

Answers

FIRST MODEL:

Well the model for the value of the house is:

$V={ \left( \frac { 11 }{ 10 } \right) }^( t )\cdot 120000$

V = Value

t = Years passed {t≥0}

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When t=0, V=120000

When t=1, V=132000

When t=2, V=145200

etc... etc...

---------------------------

Now, this model is actually curved so there is no constant rate of growth each month. We can only calculate what the rate of growth is at a particular time. If we want to find out the rate of growth at a particular time, we must differentiate the formula (model) above.

--------------------------

$V={ \left( \frac { 11 }{ 10 } \right) }^( t )\cdot 120000\n \n \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^( t )\cdot 120000 \right) } }$

$\n \n \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^( t ) \right) } } +\ln { \left( 120000 \right) } \n \n \ln { V=t\ln { \left( \frac { 11 }{ 10 } \right) } } +\ln { \left( 120000 \right) }$

$\n \n \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \n \n V\cdot \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \cdot V$

$\n \n \therefore \quad \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \cdot { \left( \frac { 11 }{ 10 } \right) }^( t )\cdot 120000$

Plug any value of (t) that is greater than 0 into the formula above to find out how quickly the investment is growing. If you want to find out how quickly the investment was growing after 1 month had passed, transform t into 1/12.

The rate of growth is being measured in years, not months. So when t=1/12, the rate of growth turns out to be 11528.42 per annum.

SECOND MODEL (What you are ultimately looking for):

$V={ \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000$

V = Value of house

t = months that have gone by {t≥0}

Formula above differentiated:

$V={ \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000\n \n \ln { V } =\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000 \right) }$

$\n \n \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } } \right) } } +\ln { \left( 120000 \right) }$

$\n \n \ln { V=\frac { t }{ 12 } } \ln { \left( \frac { 11 }{ 10 } \right) } +\ln { \left( 120000 \right) }$

$\n \n \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) }$

$\n \n V\cdot \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) } \cdot V$

$\n \n \therefore \quad \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) } \cdot { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000$

When t=1, dV/dt = 960.70 (2dp)

dV/dt in this case will measure the rate of growth monthly. As more money is accumulated, this rate of growth will rise. The rate of growth is constantly increasing as the graph of V is actually a curve. You can only find out the rate at which the house value is growing monthly at a particular time.

The formula for the lateral area of a right cone la=pirs, where r is the radius of the base and s is the slant height of the cone. which are equivalent equations?

Answers

Answer:

$\pi r √(r^2+h^2)$

Step-by-step explanation:

Given that there is a cone with radius r and slant height s.

We know that lateral surface area of the cone is

= $\pi r s$

Alternately if instead of s, h is known we can write this in terms of h also.

Consider the right triangle formed by slant height, height and radius of the cone.

Using Pythagorean theorem we have

$s^2=r^2+h^2$

Or $s=√(r^2+h^2)$

Hence lateral surface area of the cone

= $\pi r √(r^2+h^2)$

Answer:

A And D

Step-by-step explanation:

Did The Quiz and got it right

A boat travels at 15 km/hr in still water. In travelling 45 km downstream from town A to town Bit completes the journey in 75 minutes less than it takes for the return journey. At what speed does the river flow?

Answers

Let the speed of river be x km/hr down stream.

Time = Distance / speed

Since river flows downstream, speed of boat down stream = Speed of boat + speed of river = (15 + x).

Since river flows downstream, speed of boat upstream = Speed of boat - speed of river = (15 - x).

Time Upstream - Time Downstream = 75 minutes

Time Upstream = 45 / (15 - x)

Time Downstream = 45 / (15 + x)

75 minutes = 75/60 = 5/4 hours

Time Upstream - Time Downstream = 75 minutes = 5/4 hours

45 / (15 - x) - 45 / (15 + x) = 5/4 Divide both sides by 45

1 / (15 - x) - 1 / (15 + x) = (5/4)*(1/45)

1 / (15 - x) - 1 / (15 + x) = 1/36

((15 + x) - (15 -x)) / (15-x)(15+x) = 1/36

(15 +x - 15 +x) / (15-x)(15+x) = 1/36

2x / (15-x)(15+x) = 1/36

(15-x)(15+x) = 2x*36

(15-x)(15+x) = 72x

225 - x² = 72x

0 = x² + 72x -225

x² + 72x -225 = 0 This is a quadratic function, use a calculator that can solve the function, by inputting the function.

x = 3, or -75. Since we are solving for speed, we can not have negative values.

x = 3 is the only valid solution.

Speed of the river = 3 km/hr downstream.

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After the expression (x16)3/4 is simplified as much as possible, x is raised to what exponent?