**PLEASE ANSWER QUICKLY, I'LL MARK BRAINLIEST** 1/2b

Plz help and show your work

Usimov [2.4K]

The variable you wanted to find wasn’t mentioned, so I solved for all three:

5xy + n = 12

5xy = 12 - n

x = 12-n/5y

y = 12-n/5x

n = 12 - 5xy

3 0

4 years ago

The largest possible fair dice is a disdyakis triacontahedron, which has 120 sides. What is the expected value, E(X), of rolling

Luden [163]

Assuming that the 120 faces have all numbers from 1 to 120 printed on them, we have a dice that can output all numbers between 1 and 120 with the same probability of 1/120.

So, the expected value is

$\displaystyle \sum_{i=1}^{120}i\cdot p(i) = \sum_{i=1}^{120}i\cdot \dfrac{1}{120}=\dfrac{1}{120}\sum_{i=1}^{120}i=\dfrac{1}{120}\dfrac{120\cdot 121}{2}=\dfrac{121}{2}=60.5$

5 0

3 years ago

A study showed that there are 16 vehicles for every 10 households. How many vehicles are there for 107 million households?

Komok [63]

$16v........10h \\x......107.000.000h \\\\ x=\frac{16*107.000.000}{10} \\\\ x=16*10.700.000 \\\\ \boxed{x=171.200.000 \ vehicles}$

5 0

3 years ago

Read 2 more answers

What is the answer to this?

suter [353]

Answer:unpredictable

Step-by-step explanation:

8 0

3 years ago

Consider the following vector function. R(t) = 9 2 t, e9t, e−9t (a) find the unit tangent and unit normal vectors t(t) and n(t)

garik1379 [7]

The unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

<h3>What is vector?</h3>

It is defined as the quantity that has magnitude as well as direction also the vector always follows the sum triangle law.

We have vectored function:

$\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})$

Find its derivative:

$\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})$

Now its magnitude:

$\rm |R'(t) |= \sqrt{(9\sqrt{2})^2+ (9e^{9t})^2+ (-9e^{-9t})^2}$

After simplifying:

$\rm R'(t) = 9 \dfrac{e^{18t}+1}{e^{9t}}$

Now the unit tangent is:

$\rm T(u) = \dfrac{R'(t)}{|R'(t)|}$

After dividing and simplifying, we get:

$\rm T(u) = \dfrac{1}{e^{18t}+1} (\sqrt{2}e^{9t}, e^{18t}, -1)$

Now, finding the derivative of T(u), we get:

$\rm T'(u) = \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t}), 18e^{18t}, 18e^{18t})$

Now finding its magnitude:

$\rm |T'(u) |= \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t})^2+ (18e^{18t})^2+( 18e^{18t})^2)$

After simplifying, we get:

$\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}$

Now for the normal vector:

Divide T'(u) and |T'(u)|

We get:

$\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t}, \sqrt{2}e^{9t}, \sqrt{2}e^{9t})$

Thus, the unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

Learn more about the vector here:

brainly.com/question/8607618

#SPJ4

3 0

2 years ago

**PLEASE ANSWER QUICKLY, I'LL MARK BRAINLIEST** 1/2b - 15

PLEASE ANSWER QUICKLY, I'LL MARK BRAINLIEST 1/2b - 15