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3 years ago

For the functions f(x) = x^2+ 8x + 2 and g(x) = -5+9, find (f•g)(x)and (f.g)(-1).

Mathematics

1 answer:

Ne4ueva [31]3 years ago

6 0

Answer: (f·g)(x) = -5x³ - 31x² + 62x + 18

(f·g)(-1) = -70

(fog)(x) = 25x² - 130x + 155

(fog)(-1) = 310

Step-by-step explanation:

f(x) = x² + 8x + 2 g(x) = -5x + 9

(f·g)(x) = (x² + 8x + 2)(-5x + 9)

= -5x³ + 9x²

- 40x² + 72x

- 10x + 18

= -5x³ - 31x² + 62x + 18

(f·g)(-1)= -5(-1)³ - 31(-1)² + 62(-1) + 18

= -5(-1) - 31(1) - 62 + 18

= 5 - 31 - 62 + 18

= -70

****************************************************************************************

(fog)(x) = (-5x + 9)² + 8(-5x + 9) + 2

= 25x² - 90x + 81

- 40x + 72

+ 2

= 25x² - 130x + 155

(fog)(-1) = 25(-1)² - 130(-1) + 155

= 25 + 130 + 155

= 310

It wasn't clear if you wanted multiplication or composition so I solved both.

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2 years ago

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

Solution:

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

8 0

2 years ago

X^2 - 9x -10 =0 has a discriminant of: a. 121 b. 72 c. -359 d. 41 e. 136

Feliz [49]

In the standard form of quadratic
$ax^{2} +bx+c$
the discriminant is
$b^{2} -4ac$
In your quadratic, a = 1, b = -9 and c = -10
Now you need to plug these values into the expression for the discriminant.

7 0

2 years ago