All categories

3 years ago

For the function G defined by G(x) = 5x + 3, find G(2)

Mathematics

2 answers:

Arturiano [62]3 years ago

6 0

13, replace x as 2 and then solve, 5(2)+3 which is 10+3 which equals 13.

Anon25 [30]3 years ago

3 0

G(x)=5x+3

$\\ \sf\longmapsto G(2)$

$\\ \sf\longmapsto 5(2)+3$

$\\ \sf\longmapsto 10+3$

$\\ \sf\longmapsto 13$

Option c is correct

You might be interested in

Can you help me on 16, 17, 18, 19, and 20

snow_tiger [21]

Answer:

16. C 17.A 18.D 19.c

hope this helps

3 0

3 years ago

Read 2 more answers

Say that the economy is in a recession, causing vehicle prices to drop by eight percent. If your car was previously worth $25,29

qaws [65]

Answer:

$23,271

Step-by-step explanation:

i took the same test on edg

8 0

3 years ago

Read 2 more answers

Solve x2 + 14x = -40 by completing the square.

zalisa [80]

Answer: C x=-4 and -10

Step-by-step explanation:

6 0

2 years ago

Which of the following geometric series converges?

Artist 52 [7]

All three series converge, so the answer is D.

The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.

Consider a geometric sequence with the first term a and common ratio |r| < 1. Then the n-th partial sum (the sum of the first n terms) of the sequence is

$S_n=a+ar+ar^2+\cdots+ar^{n-2}+ar^{n-1}$

Multiply both sides by r :

$rS_n=ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^n$

Subtract the latter sum from the first, which eliminates all but the first and last terms:

$S_n-rS_n=a-ar^n$

Solve for $S_n$ :

$(1-r)S_n=a(1-r^n)\implies S_n=\dfrac a{1-r}-\dfrac{ar^n}{1-r}$

Then as gets arbitrarily large, the term $r^n$ will converge to 0, leaving us with

$S=\displaystyle\lim_{n\to\infty}S_n=\frac a{1-r}$

So the given series converge to

(I) -243/(1 + 1/9) = -2187/10

(II) -1.1/(1 + 1/10) = -1

(III) 27/(1 + 1/3) = 18

8 0

3 years ago

Urgent!

boyakko [2]

Step-by-step explanation:

18.

the last one.

-4.9t² + 24.5t + 117.6 = 0

the ball starts at 117.6 meters height. then, during the first seconds the thrust upwards is adding height, until finally, with more and more seconds passing, gravity will win with a vengeance and pulls the ball down faster and faster than any other force in any other direction.

19.

the general solution for a quadratic equation

y = ax² + bx + c

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = t

a = -4.9

b = 24.5

c = 117.6

t = (-24.5 ± sqrt(600.25 - -4×4.9×117.6))/-9.8 =

= (-24.5 ± sqrt(600.25 + 2304.96))/-9.8 =

= (-24.5 ± sqrt(2905.21))/-9.8 = (-24.5 ± 53.9)/-9.8

t1 = (-24.5 + 53.9)/-9.8 = -3

t2 = (-24.5 - 53.9)/-9.8 = 8

a negative time does not make any sense, so the real solution is : the ball reached the ground after 8 seconds.

20.

now the equation is

-4.9t² + 24.5t + 117.6 = 49

but this is quickly transformed again into a "= 0" equation :

-4.9t² + 24.5t + 68.6 = 0

t = (-24.5 ± sqrt(600.25 - -4×4.9×68.6))/-9.8 =

= (-24.5 ± sqrt(600.25 + 1344.56))/-9.8 =

= (-24.5 ± sqrt(1944.81))/-9.8 = (-24.5 ± 44.1)/-9.8

t1 = (-24.5 + 44.1)/-9.8 = -2

t2 = (-24.5 - 44.1)/-9.8 = 7

again, negative time does not make sense.

the ball will be at 49 m height after 7 seconds.

5 0

2 years ago