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Marat540 [252]
3 years ago
13

Solve for x: −2x + 5 < 7

Mathematics
1 answer:
hoa [83]3 years ago
7 0
-2x + 5 < 7

-2x + 5 < 7
<u>        -5     -5   </u>       deduct 5 from both sides
-2x       < 2
<u>÷ -2        ÷ -2  </u>       divide both sides by negative 2. Because of the division 
 x          > -1          using a negative number, the sign is then reversed.
                             from < it becomes >.

The value of x should be greater than -1. It can be 0, 1, 2, so on... 

To check: Revert back to the original sign which is <.

x = 1
-
2x + 5 < 7
-2(1) + 5 < 7
-2 + 5 < 7
3 < 7
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A CD usually sells for $14.00. If the CD is 20% off, and sales tax is 7%, what is the total price of the CD, including tax?
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Answer:

the total price of the CD including tax is $11.984

Step-by-step explanation:

The computation of the total price of the CD including tax is shown below:

The Selling price of the CD is $14

CD is 20% off

And, the sales tax rate is 7%

So, the total price of the CD is

= ($14 - $14 × 20%) × (1.07)

= $11.2 × 1.07

= $11.984

Hence, the total price of the CD including tax is $11.984

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PEASE HELP REALLY QUICK
8_murik_8 [283]

Answer:

x = 1

Step-by-step explanation:

13x - 12 = 7x -6

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Remember the sign opens to the bigger number and the line underneath is only when that value is included.
8 0
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Givea)Possible number of positive real rootsb)Possible number of negative real rootsc)Possible rational roolsd)Find the roots
CaHeK987 [17]

The function is given to be:

x^3-2x^2-3x+6

QUESTION A

We can use Descartes' Rule of Signs to check the positive real roots of a polynomial.

The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number.

If we have:

f(x)=x^3-2x^2-3x+6

The coefficients are: +1, -2, -3, +6.

We can see that there are only 2 sign changes; from the first to the second term, and from the third to the fourth term.

Therefore, there are 2 or 0 positive real roots.

QUESTION B

To find the number of negative real roots, evaluate f(-x) and check for sign changes:

\begin{gathered} f(-x)=(-x)^3-2(-x)^2-3(-x)+6 \\ f(-x)=-x-2x^2+3x+6 \end{gathered}

The coefficients are: -1, -2, +3, +6.

We can see that there is only one sign change; from the second term to the third term.

Therefore, there is 1 negative real root.

QUESTION C

To check the possible rational roots, we can use the Rational Root Theorem since all the coefficients are integers.

The Rational Root Theorem states that if the polynomial:

P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0

has any rational roots, they must be in the form:

\Rightarrow\pm\mleft\lbrace\frac{factors\text{ of }a_0}{factors\text{ of }a_n}\mright\rbrace

From the polynomial, the trailing coefficient is 6:

a_o=6

Factors of 6:

\Rightarrow\pm1,\pm2,\pm3,\pm6

The leading coefficient is 1:

a_n=1

Factors of 1:

\Rightarrow\pm1_{}

Write in the form

\Rightarrow\mleft\lbrace\frac{a_o}{a_n}\mright\rbrace

Therefore,

\Rightarrow\pm(\frac{1}{1}),\pm(\frac{2}{1}),\pm(\frac{3}{1}),\pm(\frac{6}{1})

Therefore, the possible rational roots are:

\Rightarrow\pm1,\pm2,\pm3,\pm6

QUESTION D

We can use a graph to check the roots of the polynomial. The graph is shown below:

The roots of the polynomial refer to the points when the graph intersects the x-axis.

Therefore, the roots of the polynomial are:

x=-1.732,x=1.732,x=2

7 0
2 years ago
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