All categories

3 years ago

If f(x) = x2 – 2x and g(x) = 6x + 4, for which value of x does (f + g)(x) = 0?

Mathematics

2 answers:

Sholpan [36]3 years ago

7 0

Answer: The required value of x is -2.

Step-by-step explanation: We are given the expressions for two functions f(x) and g(x) as follows :

$f(x)=x^2-2x,~~~~~~g(x)=6x+4.$

We are to find the values of x for which (f + g)(x) = 0.

We know that

for any two functions p(x) and q(x),

$(p+q)(x)=p(x)+q(x).$

We have

$(f+g)(x)=0\\\\\Rightarrow f(x)+g(x)=0\\\\\Rightarrow x^2-2x+6x+4=0\\\\\Rightarrow x^2+4x+4=0\\\\\Rightarrow x^2+\times2x+2^2=0\\\\\Rightarrow (x+2)^2=0\\\\\Rightarrow x+2=0,~~x+2=0\\\\\Rightarrow x=-2.$

Thus, the required value of x is -2.

alekssr [168]3 years ago

5 0

G(f(x)) = 2(x^2 -1) + 3

= 2x^2 + 1

subsitute 2

= 9

Hence, the answer is 9

You might be interested in

Can someone please help me with this :(

konstantin123 [22]

Answer:

(it will take super long for me to answer this)

Step-by-step explanation:

Use desmos graphing calculator, put in your f(x)/y values and you will get all of the answers you need. Hope this is helpful :)

5 0

2 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

2 years ago

Find the value of x. The diagram is not to scale.

DiKsa [7]

Sum of interior angles in a pentagon=540°

148+112+2x+2x+10+90=540

360+4x=540

4x=540-360

4x=180

x=180/4

x=45°

5 0

2 years ago

If (x+30) and (3x+10)are co interior angle.Find the value of x

Amiraneli [1.4K]

Answer:19

Step-by-step explanation:

3x3=9 then add by 10 giving u the sum of 19

3 0

3 years ago

Read 2 more answers

ASAP I NEED HELP NOWW

marshall27 [118]

27 x 1.25= 33.75

5 x 1.99=9.95

1 x 1.50= 1.50

add that up =45.2

/26 students = 1.73 dollars

4 0

2 years ago

Read 2 more answers