All categories

3 years ago

What is the root of the polynomial equation *(x-2)(x+3) - 18? Use a graphing calculator and a system of

Mathematics

2 answers:

zysi [14]3 years ago

5 0

Answer:

Start by using your first factor, 1. Substitute "1" for each "x" in the equation: (1)3 - 4(1)2 - 7(1) + 10 = 0.

This gives you: 1 - 4 - 7 + 10 = 0.

Because 0 = 0 is a true statement, you know that x = 1 is a solution.

Step-by-step explanation:

Start by using your first factor, 1. Substitute "1" for each "x" in the equation: (1)3 - 4(1)2 - 7(1) + 10 = 0.

This gives you: 1 - 4 - 7 + 10 = 0.

Because 0 = 0 is a true statement, you know that x = 1 is a solution.

Katarina [22]3 years ago

5 0

Answer:

The answer is 3 :)

Step-by-step explanation:

EDGE2020

You might be interested in

Which is the largest? 76%,4/5, .77

lord [1]

4/5 is the same as 80% or .80, so it's larger than 76% and .77

5 0

4 years ago

Consider the inequality x> 7.

kramer

Answer:

x is greater than 7, so x can be 8 or more.

4 0

3 years ago

Read 2 more answers

Find a second solution y2(x) of<br> x^2y"-3xy'+5y=0; y1=x^2cos(lnx)

rosijanka [135]

We can try reduction order and look for a solution $y_2(x)=y_1(x)v(x)$ . Then

$y_2=y_1v\implies{y_2}'=y_1v'+{y_1}'v\implies{y_2}''=y_1v''+2{y_1}'v+{y_1}''v$

Substituting these into the ODE gives

$x^2(y_1v''+2{y_1}'v+{y_1}''v)-3x(y_1v'+{y_1}'v)+5y_1v=0$

$x^2y_1v''+(2x^2{y_1}'-3xy_1)v'+(x^2{y_1}''-3x{y_1}'+5y_1)v=0$

$x^4\cos(\ln x)v''+x^3(\cos(\ln x)-2\sin(\ln x))v'=0$

which leaves us with an ODE linear in $w(x)=v'(x)$ :

$x^4\cos(\ln x)w'+x^3(\cos(\ln x)-2\sin(\ln x))w=0$

This ODE is separable; divide both sides by the coefficient of $w'(x)$ and separate the variables to get

$w'+\dfrac{\cos(\ln x)-2\sin(\ln x)}{x\cos(\ln x)}w=0$

$\dfrac{w'}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}$

$\dfrac{\mathrm dw}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}\,\mathrm dx$

Integrate both sides; on the right, substitute $u=\ln x$ so that $\mathrm du=\dfrac{\mathrm dx}x$ .

$\ln|w|=\displaystyle\int\frac{2\sin u-\cos u}{\cos u}\,\mathrm du=\int(2\tan u-1)\,\mathrm du$

Now solve for $w(u)$ ,

$\ln|w|=-2\ln(\cos u)-u+C$

$w=e^{-2\ln(\cos u)-u+C}$

$w=Ce^{-u}\sec^2u$

then for $w(x)$ ,

$w=Ce^{-\ln x}\sec^2(-\ln x)$

$w=C\dfrac{\sec^2(\ln x)}x$

Solve for $v(x)$ by integrating both sides.

$v=\displaystyle C_1\int\frac{\sec^2(\ln x)}x\,\mathrm dx$

Substitute $u=\ln x$ again and solve for $v(u)$ :

$v=\displaystyle C_1\int\sec^2u\,\mathrm du$

$v=C_1\tan u+C_2$

then for $v(x)$ ,

$v=C_1\tan(\ln x)+C_2$

So the second solution would be

$y_2=x^2\cos(\ln x)(C_1\tan(\ln x)+C_2)$

$y_2=C_1x^2\sin(\ln x)+C_2x^2\cos(\ln x)$

$y_1(x)$ already accounts for the second term of the solution above, so we end up with

$\boxed{y_2=x^2\sin(\ln x)}$

as the second independent solution.

6 0

4 years ago

Will you help me i am have a lot of trouble on this on

vodka [1.7K]

Answer:

A = 18 cm²

Step-by-step explanation:

The volume (V) of a prism is calculated as

V = area of base (A) × height

Here V = 216 cm³ and h = 12 cm , then

A × 12 = 216 ( divide both sides by 12 )

A = 18 cm² ← area of base

7 0

3 years ago

Identify all sets of numbers to which 2.55 belong

Sloan [31]

Answer: Real number, Rational number.

Step-by-step explanation:

To solve this exercise you must know the following definitions:

1. A Rational number is a number that you can write as a fraction, whose numerator and denominator are integers.

2. By definition, the Real numbers contain Natural numbers, Whole numbers, Integers, Rational numbers and Irrational numbers.

3. Then, the number 2.55 can be written as a fraction: $\frac{51}{20}$ , therefore, it is a Real number and a Rational Number.

7 0

3 years ago