How would you express 325.4 miles on 17.3 gallons as a unit rate

1 answer:

6 0

I think the answer is 18.8 Miles/gallons. 325.4 miles / 17.3 gallons = 18.78 miles/gallons, round it to 18.8 miles/gallons.

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Question:

djyliett [7]

I think just the "possible" ones, without fully factoring, are based on factors of the last whole number-- +20.
So factors of 20: 1, 2, 4, 5, 10, 20

8 0

3 years ago

How can you use factors to write a number as a product?

vekshin1

You would have to multiple the two factors to get the number as a product.

8 0

3 years ago

I don't understand this at all

suter [353]

Answer:same :(

Step-by-step explanation:

6 0

3 years ago

Brandon wants to test drive a Lamborghini car and he needs the key. Each key ring has 6 keys. If there are 9 key rings, how many

Harrizon [31]

Answer:

54 keys

Step-by-step explanation:

If there are 6 keys on each of the nine key rings, you times 6 by 9 to get 54.

6 0

3 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.

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