Solve the logarithmic equation. Round to the nearest ten-thousandth if necessary. log(x+9)-log x=3

These tables represent an exponential function.Find the average rate of change for the interval from x=8 to x=9

GuDViN [60]

Answer:

Option D. 13122 is the answer.

Step-by-step explanation:

As we can see from the table having interval and average rate of change, figures under average rate of change are forming a geometric sequence.

Sequence is 2, 6, 18 , 54, 162, 486.

and we have to find the average rate of change from x = 8 to x = 9, means we have to find 9th term of the given sequence.

Now we know that explicit formula of the sequence can be written as $T_{n}=ar^{n-1}$

where Tn is the nth term of the sequence.

a = first term

r = common ratio

n = number of the term

Now from this explicit formula we can find the 9th term of the sequence.

From the given table

a = 2, r = 3, n = 9

$T_{9}=2.3^{9-1}=2.3^{8}$

T9 = 13122

Therefore Option D. 13122 will be the answer.

6 0

2 years ago

100

Tpy6a [65]

This is the most easier question anyone has asked yet. kill me

8 0

2 years ago

Which is the correct input-output table for the function f(x) = 7 – 4.5x?

Naddika [18.5K]

Step 1: Divide both sides by x.fxx=−4.5x+7xf=−4.5x+7xAnswer:f=−4.5x+7x

3 0

2 years ago

Read 2 more answers

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

2 years ago

If (a, -5) is a solution to the equation 3a = -2B - 7, what is a?

Andrew [12]

I think what you need to do is make B = -5
3a = -2(-5) - 7
3a = 10 - 7
3a = 3
a = 1

3 0

3 years ago