Solve open parentheses square root of 6 close parentheses to the 8 x power = 216x−3. (1 point)

A local school has was evaluated the the Education Department and received the

Angelina_Jolie [31]

Answer: 79

Step-by-step explanation:

Given the following :

Graduation Rates:

Grade:82

Weight: 50%

Passing Rates:

Grade: 75

Weight: 30%

Enrollment Numbers:

Grade: 60

Weight: 10%

Career readiness:

Grade: 95

Weight: 10%

Overall score : Σ(grade × weight)

(82 × 50%) + (75 × 30%) + (60 × 10%) + (95 × 10%)

= (41 + 22.5 + 6 + 9.5)

= 79

Hence, overall score is 79

5 0

3 years ago

Please help me out solve it and tell me how you got that answer so I can understand please!

NikAS [45]

The solution for s in the given equation is $s = C-nD$

The question seems to be incomplete

Here is the complete question:

Solve for s in this equation $D= \frac{C-s}{n}$ Depreciation.

To solve for s, that means we should make s the subject of the equation

From the given equation,

$D= \frac{C-s}{n}$

To solve for s, first multiply both sides by n to clear the fraction

We get

$n\times D= n \times \frac{C-s}{n}$

Then,

$nD = C - s$

Now, add s to both sides

$nD + s= C - s+s$

$nD+s = C$

Then, subtract nD from both sides

$nD-nD+s = C-nD$

∴ $s = C-nD$

Hence, the solution for s in the given equation is $s = C-nD$

Learn more here: brainly.com/question/21406377

6 0

3 years ago

Find the 75th term of the arithmetic sequence -1, 15, 31, ...−1,15,31,...

Readme [11.4K]

Answer:

1183

Step-by-step explanation:

Common difference:

d = 31 - 15 = 16

nth term:

$a_n = a_1 + (n - 1)d$

Here, $a_1 = -1$

$a_{75} = -1 + (75 - 1)(16)$

$a_{75} = 1183$

3 0

3 years ago

Find the perimeter of the triangle. Recall that the perimeter of a figure is the distance around a figure.

Svet_ta [14]

Answer:

1 inch

Step-by-step explanation:

The perimeter is the sum of the lengths of the sides:

Perimeter = 3/23 + 9/23 + 11/23 = (3 + 9 + 11)/23 = 23/23 = 1

8 0

2 years ago

PLEASE HELP ASAP! I don’t recall how to do this!

MakcuM [25]

Answer:

Step-by-step explanation:

For a. we start by dividing both sides by 200:

$(1.05)^x=1.885$

In order to solve for x, we have to get it out from its position of an exponent. Do that by taking the natural log of both sides:

$ln(1.05)^x=ln(1.885)$

Applying the power rule for logs lets us now bring down the x in front of the ln:

x * ln(1.05) = ln(1.885)

Now we can divide both sides by ln(1.05) to solve for x:

$x=\frac{ln(1.885)}{ln(1.05)}$

Do this on your calculator to find that

x = 12.99294297

For b. we will first apply the rule for "undoing" the addition of logs by multipllying:

$ln(x*x^2)=5$

Simplifying gives you

$ln(x^3)=5$

Applying the power rule allows us to bring down the 3 in front of the ln:

3 * ln(x) = 5

Now we can divide both sides by 3 to get

$ln(x)=\frac{5}{3}$

Take the inverse ln by raising each side to e:

$e^{ln(x)}=e^{\frac{5}{3}}$

The "e" and the ln on the left undo each other, leaving you with just x; and raising e to the power or 5/3 gives you that

x = 5.29449005

For c. begin by dividing both sides by 20 to get:

$\frac{1}{2}=e^{.1x}$

"Undo" that e by taking the ln of both sides:

$ln(.5)=ln(e^{.1x})$

When the ln and the e undo each other on the right you're left with just .1x; on the left we have, from our calculators:

-.6931471806 = .1x

x = -6.931471806

Question d. is a bit more complicated than the others. Begin by turning the base of 4 into a base of 2 so they are "like" in a sense:

$(2^2)^x-6(2)^x=-8$

Now we will bring over the -8 by adding:

$(2^2)^x-6(2)^x+8=0$

We can turn this into a quadratic of sorts and factor it, but we have to use a u substitution. Let's let $u=2^x$

When we do that, we can rewrite the polynomial as

$u^2-6u+8=0$

This factors very nicely into u = 4 and u = 2

But don't forget the substitution that we made earlier to make this easy to factor. Now we have to put it back in:

$2^x=4,2^x=2$

For the first solution, we will change the base of 4 into a 2 again like we did in the beginning:

$2^2=2^x$

Now that the bases are the same, we can say that

x = 2

For the second solution, we will raise the 2 on the right to a power of 1 to get:

$2^x=2^1$

Now that the bases are the same, we can say that

x = 1

5 0

3 years ago