If (2x + 4) - (x + 3) = 15, then x =

Two terms in a geometric sequence are a3=729 and a4=243.

lesantik [10]

<h2>The first term of the given sequence (a) = 6561</h2>

Step-by-step explanation:

Let the first term = a and common difference = d

Given,

$a_{3}$ = 729 and $a_{4}$ = 243

To find, the first term of the given sequence (a) = ?

We know that,

The nth term of a G.P.

$a_{n} =ar^{n-1}$

The 3rd term of a G.P.

$a_{3} =ar^{3-1}$

⇒ $ar^{2}$ = 729 ..............(1)

The 4th term of a G.P.

$a_{4} =ar^{4-1}$

⇒ $ar^{3}$ = 243 ..............(2)

Dividing equation (2) by (1), we get

$\dfrac{ar^{3}}{ar^{2}}$ = $\dfrac{243}{729}$

⇒ $r=\dfrac{1}{3}$

Put $r=\dfrac{1}{3}$ in equation (1), we get

$a(\dfrac{1}{3})^{2}$ = 729

⇒ $a(\dfrac{1}{9})$ = 729

⇒ a = 9 × 729 = 6561

∴ The first term of the given sequence (a) = 6561

8 0

3 years ago

Jennifer is flying a kite. She is looking up at the kite at an angle of elevation of 39º. If the hand that is holding the kite s

inn [45]

See attachment for answer.

5 0

2 years ago

I’ll put you as brainliest

kari74 [83]

The law of an object moving with constant acceleration is

$s(t)=s_0+v_0t+\dfrac{1}{2}at^2$

Where $s$ is space, $t$ is time, $s_0$ is the initial position, $v_0$ is the initial velocity and $a$ is the acceleration.

In this case, if we choose a reference grid with the vertical axis pointing upwards, the acceleration of gravity will point downwards (and thus be negative). The initial position is zero, because the rocket is on the ground, and the initial velocity is 100 (positive because pointing upwards).

So, its law is

$h(t)=100t-\dfrac{1}{2}gt^2$

(I changed $s$ for $h$ since the rocket is moving vertically, so its position is actually its height. Also, g is the acceleration due to gravity).

The rocket hits the ground if its height is zero, so if we set $h=0$ we have

$0=100t-\dfrac{1}{2}gt^2 \iff t(100-\dfrac{1}{2}gt)=0$

Solving for t, we have either t=0, or

$100-\dfrac{1}{2}gt=0 \iff 100=\dfrac{1}{2}gt \iff 200=gt \iff t=\dfrac{200}{g}$

The solution t=0 means that at the beginning the rocket is on the ground. So, we're interested in the other solution. Considering that g is about 32.2 feet/s^2, we have

$t=\dfrac{200}{g}\approx \dfrac{200}{32.2}\approx 6.21$

3 0

3 years ago

3x+7=8x+7<br><br> a. infinite solutions<br> b. one solution<br> c. no solution

kykrilka [37]

Answer:

The above has one solution

Hope this helps

4 0

3 years ago

Give the following equations determine if the lines are parallel perpendicular or neither

GaryK [48]

In order to determine whether the equations are parallel, perpendicular, or neither, let's simply each equation into a slope-intercept form or basically, solve for y.

Let's start with the first equation.