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Dmitry_Shevchenko [17]
3 years ago
11

Which graph shows the solution to this system of inequalities? y>-1/3x+1 y>2x-3

Mathematics
1 answer:
Mademuasel [1]3 years ago
6 0

Given:

The system of inequalities is:

y>-\dfrac{1}{3}x+1

y>2x-3

To find:

The graph of the given system of inequalities.

Solution:

We have,

y>-\dfrac{1}{3}x+1

y>2x-3

The related equations are:

y=-\dfrac{1}{3}x+1

y=2x-3

Table of values for the given equations is:

    x                   y=-\dfrac{1}{3}x+1             y=2x-3

   0                             1                              -3

   3                             0                              3

Plot (0,1) and (3,0) and connect them by a straight line to get the graph of y=-\dfrac{1}{3}x+1.

Similarly, plot (0,-3) and (3,3) and connect them by a straight line to get the graph of y=2x-3.

The signs of both inequalities are ">". So, the boundary line is a dotted line and the shaded region or each inequality lie above the boundary line.

Therefore, the graph of the given system of inequalities is shown below.

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Elena L [17]
Remark
If there are 5 distinct zeros that means either that the x axis is crossed the x axis 5 different places or touched the x axis in 1 place out the 5. Touching in one place means that an even number of roots are the same. 

So let's go through all of them to get an answer of 5.

A has 4 x intercepts. It is not the right answer. We need 5.
B has 4 x intercepts. It is not the right answer. We need 5.
C has 6 x intercepts. Not the one we want.
D has 5 x distinct zeros. The wording is a bit tricky. It does not matter than one of them just touches the x axis. There could be an even number of distinct zeros there, but it only counts as one root.

An example of such a graph is f(x)=\left(\frac{1}{10}(x+2)(x+1.5)(x+1)(x-2)^4(x-3)\right)
 
Answer D <<<<<


6 0
4 years ago
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How much less would a 23-year-old female pay for a $25,000 policy of 20 year life insurance (@ $2.90 per $1000) than straight li
mojhsa [17]

<u>Answer</u>:

The female would pay  $322.00 less for a policy of $25,000

<u>Step-by-step explanation:</u>

Since we have given that

Amount for policy = $25000

If she opt for 20 year life insurance at $2.90 per $1000.

so, her amount of premium becomes

25000\times \frac{2.9}{1000}

=$72.50

If she opt for straight life insurance at $15.78 per $1000,

Then, her amount of premium becomes

25000 \times \frac{15.78}{1000}

= $394.50

Difference between them is given by

$394.50-$72.5 = $322.00

5 0
4 years ago
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In a two-step equation, you should take care of multiplication and division before addition and subtraction.
Vladimir79 [104]
I’m not sure if this is a question, but yes, i like to use “PEMDAS”.
Parentheses, exponents, multiply, divide, add, subtract.
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4 years ago
How to find inequalities
geniusboy [140]

Answer:

Step 1 Eliminate fractions by multiplying all terms by the least common denominator of all fractions.

Step 2 Simplify by combining like terms on each side of the inequality.

Step 3 Add or subtract quantities to obtain the unknown on one side and the numbers on the other.

Hope that helped happy holidays!

6 0
3 years ago
Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A
Vlad [161]

Answer:

y(x)=sin(2x)+2cos(2x)

Step-by-step explanation:

y''+4y=0

This is a homogeneous linear equation. So, assume a solution will be proportional to:

e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda

Now, substitute y(x)=e^{\lambda x} into the differential equation:

\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0

Using the characteristic equation:

\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0

Factor out e^{\lambda x}

e^{\lambda x}(\lambda ^2 +4) =0

Where:

e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda

Therefore the zeros must come from the polynomial:

\lambda^2+4 =0

Solving for \lambda:

\lambda =\pm2i

These roots give the next solutions:

y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}

Where c_1 and c_2 are arbitrary constants. Now, the general solution is the sum of the previous solutions:

y(x)=c_1 e^{2ix} +c_2 e^{-2ix}

Using Euler's identity:

e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)

y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\

Redefine:

i(c_1-c_2)=c_1\\\\c_1+c_2=c_2

Since these are arbitrary constants

y(x)=c_1sin(2x)+c_2cos(2x)

Now, let's find its derivative in order to find c_1 and c_2

y'(x)=2c_1 cos(2x)-2c_2sin(2x)

Evaluating    y(0)=2 :

y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2

Evaluating     y'(0)=2 :

y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1

Finally, the solution is given by:

y(x)=sin(2x)+2cos(2x)

5 0
4 years ago
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