Multiply [5 0 3 -5]•[2 -1 2 -2

9

Kamila [148]

Hope this helps
1.$40
2.$42.80
3.$51.36

5 0

2 years ago

A 5-pound bag of sugar costs $3.90. What is the unit rate?

Dimas [21]

0.78 per pound is unit rate

Since 5lbs is $3.90, 0.78 is the cost of one pound.

3.90 divided by 5 is 0.78

7 0

2 years ago

If Log 20 = 1.301 what is the value of Logsub100 20?

Charra [1.4K]

Answer:

$log_{100}20 = 0.6505$

Step-by-step explanation:

log20 = 1.301 (when you see only "log" with no base, it is taken as a base 10)

This basically means:

$10^{1.301} = 20$ (This is the log in exponential form)

Since we don't know what $log_{100}20$ is equal to, we will say $log_{100}20$ = x

So to solve for $log_{100}20$ = x, you do the same thing. (convert to exponential form)

$100^{x} =20$

Now you will notice that both of these equations are equal to 20.

Since 20 = 20,

we can say $100^{x}$ = $10^{1.301}$

Another way of saying 100, is $10^{2}$ (make the bases the same)

Now we get $10^{2x} = 10^{1.301}$

(we get 2x because an exponent to the power of an exponent ( $2^{x}$ ) is the same as 2 * x)

Because you have the same base, you can just ignore the 10s and focus on the exponents. So you get:

2x = 1.301

x = 0.6505

6 0

2 years ago

Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$