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2 years ago

Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

Mathematics

2 answers:

zmey [24]2 years ago

7 0

Answer by JKismyhusbandbae:

$\frac{x}{4}=20\\\mathrm{Multiply\:both\:sides\:by\:}4\\\frac{4x}{4}=20\cdot \:4\\Simplify\\x=80$

balandron [24]2 years ago

4 0

remember that you can do anything to an equaiton as long as you do it to both sides

x/4=20

multiiply both sides by 4

4(x/4)=20*4

(x)(4)(1/4)=80

(x)(4/4)=80

x(1)=80

x=80

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Rewrite equation in slope intercept form 3x+5y=12

pochemuha

Answer:

y= -3/5x + 12/5

Step-by-step explanation:

6 0

2 years ago

What is the next term in the sequence below? -324,108,-36,12,...

liq [111]

This is a geometric sequence with a common ratio of -1/3 and an initial term of -324. Any geometric sequence can be expressed as:
a(n)=ar^(n-1), in this case a=-324 and r=-1/3 so
a(n)=-324(-1/3)^(n-1) so the 5th term will be
a(5)=-324(-1/3)^4
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5 0

3 years ago

Read 2 more answers

Four friends plan to rent a car to go on a camping trip. They will drive a total of 450 miles. The cost of gasoline is $2.20 pe

Vlada [557]

Answer:

t = (m / g) * p

Step-by-step explanation:

Since we are given a variable for every part of the equation that we need we can go ahead and create the following equation where t equals the total cost

t = (m / g) * p

In this equation, we need to first divide the total number of miles by the number of miles that the car gets per gallon. This would give us the total number of gallons that the car would use during the entire trip. We then multiply the number of gallons by the price per gallon to get the final cost of the entire trip.

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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.

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