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3 years ago

For y = 3x - 2x^2 +5x^3 show how to find the value for y, when x = 3. Show your work

2 answers:

8 0

$y \: = \: f(x) \: = \: 3x \: - \: 2 {x}^{2} \: + \: 5 {x}^{3}$
Here, y is a cubic function of x.
When x = 3,
$y \: = \: f(3) \: = \: 3(3) \: - \: 2 {(3)}^{2} \: + \: 5 {(3)}^{3}$
$y \: = \: 9 \: - \: 18 \: + \: 135$
$y \: = \: f(3) \: = \: 126$
Hence, when x = 3, y = 126.

liq [111]3 years ago

7 0

Y=3(3)-2(3)^2+5(3)^3
y=9-2(3)^2+5(3)^3
y=9-6^2+5(3)^3
y=9-36+5(3)^3
y=9-36+15^3
y=9-36+3,375
y=-27+3,375
y=3,348

Just do the steps as I showed you. Just go little by little for each thing so that you don't mess up our get super confused.

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F(x) = 3x^3 - 2x^2 - 11x + 10; (x + 2) I need to factor and find the zeros.

Ratling [72]

Answer:

$\textsf{Factored function}: \quad f(x)=(x+2)(3x-5)(x-1)$

$\textsf{Zeros}: \quad x=-2, \quad x=\dfrac{5}{3},\quad x=1$

Step-by-step explanation:

Given function:

$f(x)=3x^3-2x^2-11x+10$

If (x + 2) is a factor then:

$\implies f(x)=(x+2)(ax^2+bx+c)$

Expand:

$\implies f(x)=ax^3+bx^2+cx+2ax^2+2bx+2c$

$\implies f(x)=ax^3+2ax^2+bx^2+2bx+cx+2c$

$\implies f(x)=ax^3+(2a+b)x^2+(2b+c)x+2c$

To find a, compare the coefficients of x³:

$\implies a=3$

To find b, substitute the found value of a into the coefficient for x² and compare:

$\implies 2a+b=-2$

$\implies 2(3)+b=-2$

$\implies b=-8$

To find c, compare the constants:

$\implies 2c=10$

$\implies c=5$

Therefore:

$\implies f(x)=(x+2)(3x^2-8x+5)$

Now factor (3x²-8x+5):

$\implies 3x^2-3x-5x+5$

$\implies 3x(x-1)-5(x-1)$

$\implies (3x-5)(x-1)$

Therefore the factored function is:

$\implies f(x)=(x+2)(3x-5)(x-1)$

Zero Product Property

$\boxed{\text{If \; $a \cdot b = 0$ \; then either \; $a = 0$ \;or \;$b = 0$ (or both)}.}$

To find the zeros, set each factor equal to zero and solve for x:

$\implies x+2=0 \implies x=-2$

$\implies 3x-5=0 \implies x=\dfrac{5}{3}$

$\implies x-1=0 \implies x=1$

Therefore, the zeros of the function are:

$x=-2, \quad x=\dfrac{5}{3},\quad x=1$

3 0

1 year ago

Grace and her father spent 4 1/2 hours over the weekend restoring their fishing boat. This time makes up 6% of the

solong [7]

$\frac{9}{2}$ multiplied by $\frac{6}{100}$
= 75 hours
4 and a half hour repairs 6% of the entire boat so it is
4 1/2 multiplied by 6/100's inverse number

6 0

3 years ago

Read 2 more answers

Consider that x=5 and y=7. Is the sum of x + y rational, irrational, or neither irrational nor rational!

nasty-shy [4]

The sum of y+x, or 5+7, is rational. This is because each integer(non fraction/decimal)is rational, it is also rational.

5 0

3 years ago

Samir is trying to decide between two checking account plans. After researching plans at two banks, he finds that Unity Bank off

GREYUIT [131]

Answer:

The monthly compounded interest rate of 0.14% of Unity Bank is a better plan.

Step-by-step explanation:

Step 1 : Write the formula for calculating a monthly compound interest rate and for calculating an annually compounded interest rate.

Monthly compound interest rate = P(1+r/n)^nxt

n=12 (number of months in a year)

t=1 (number of years)

Annually compound interest rate = P(1+r/n)^nxt

n=1 (because it is for 1 year only)

t=1 (number of years=1)

Step 2 : Lets assume that P is $100 in both banks and time is 1 year.

Step 3 : Lets substitute the values to find out which one is better.

Monthly compound interest rate = P(1+r/n)^nxt

Monthly compound interest rate = 100(1+0.14/12)^12x1

Monthly compound interest rate = 114.93

114.93 - 100 = $14.93 per month

14.93 x 12 = $179.16 for 12 months or 1 year

Annually compound interest rate = P(1+r/n)^nxt

Annually compound interest rate = 100(1+1.6/1)^1x1

Annually compound interest rate = 260

260-100 = $160 for 1 year

Therefore, the monthly compounded interest rate of 0.14% of Unity Bank is a better plan.

8 0

3 years ago

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Find the six trig function values of the angle 240*Show all work, do not use calculator

-BARSIC- [3]

Solution:

Given:

$240^0$

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

$sin240^0=sin(180+60)$

Using the trigonometric identity;

$sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny$

Hence,

$\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}$

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

$cos240^0=cos(180+60)$

Using the trigonometric identity;

$cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny$

Hence,

$\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\ \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}$

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

$tan240^0=tan(180+60)$

Using the trigonometric identity;

$tan(180+x)=tan\text{ }x$

Hence,

$\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\ \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}$

To get cosec 240 degrees:

$\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}$

To get sec 240 degrees:

$\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\ \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}$

To get cot 240 degrees:

$\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\ \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}$

5 0

1 year ago