All categories

2 years ago

Which graph shows the following system of equations and its solutions? -0.1x-0.3y=1.2 and 0.2x-0.5y=2

Mathematics

1 answer:

Vera_Pavlovna [14]2 years ago

7 0

Answer:

Option b is the correct answer as both the equations are true for given solution.

Step-by-step explanation:

Given equations are:

-0.1x-0.3y=1.2

0.2x-0.5y=2

We can observe each graph and find the point that is the solution and put the point in the equations to know if that point is the solution

For option A:

(0,4)

Putting x=0 and y = 4 in both equations

$-0.1(0)-0.3(4)=1.2\\0-1.2 =1.2\\-1.2 \neq 1.2\\0.2(0)-0.5(4)=2\\0-2 = 2\\-2 \neq 2$

This is not the correct answer as both equations are not true with this solution

For Option B:

(0,-4)

Putting x = 0 and y = -4 in both equations

$-0.1(0)-0.3(-4)=1.2\\0+1.2 = 1.2\\1.2 = 1.2\\0.2x-0.5y=2\\0.2(0)-0.5(-4) = 2\\2 = 2$

Both equations are true for (0,-4) hence it is the solution of the system.

For Option C:

(4,0)

$-0.1(4)-0.3(0)=1.2\\-0.4-0 = 1.2\\-0.4 \neq 1.2\\0.2x-0.5y=2\\0.2(4)-0.5(0) = 2\\0.8 \neq 2$

Not true for both equations

Hence,

Option b is the correct answer.

You might be interested in

PLS HELP :(

faust18 [17]

Answer: 47.1

Step-by-step explanation:

1/3πr^2h

=1/3×π×32×5

=15π

= 47.123889803847 feet3

4 0

2 years ago

Read 2 more answers

Which segment is longer, please explain

Molodets [167]

Answer:

Blue, if you look, they both are close, but due to the pathagrium theory blue is longer.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

2 years ago

If c(x)=3x-1 and k(x)=x^2-2x+1, what is the value of (c°k)(5)

Verdich [7]

Is the equation supposed to be like this

(c o f)(5)?

7 0

3 years ago

Sanji paid 25 for two scarves which were different prices when she got home she could not find the receipt she remember that one

laila [671]

Answer:

The answer is x = 14 the more expensive scarf

Step-by-step explanation:

x + y = 25

3 + y + y = 25 (x = 3 + y)

3 + 2y = 25

2y = 25 - 3

2y = 22

y = 22/2

y = 11

x = 3 + y

x = 3 + 11

x = 14 (the more expensive scarf)

(x,y) ; (14,11)

5 0

3 years ago