All categories

3 years ago

Solve each system using substitution. 4. y = 1/x + 4 x-y=8

Mathematics

1 answer:

liubo4ka [24]3 years ago

7 0

Answer:

$y=\frac{33}{8}$

Step-by-step explanation:

$y=\frac{1}{8} +4$

Use the slope-intercept form to find the slope and y-intercept.

Slope:

y-intercept:

$\frac{33}{8}$

To get the answer i used a graph

then i tried just evaluating the equation

Reduce the expression, if possible, by cancelling the common factors.

$y=\frac{33}{8}$

I got the same answer so i believe this should be your Answer

<u>Hope this helps</u>

You might be interested in

The area of a triangle is 150 cm- and its sides are in the ratio 3: 4:5. What is its perimeter?

dmitriy555 [2]

Hey, your answer is d) 60cm

6 0

2 years ago

what's the error? MAX said that 36,594,145 is less than 5,980,251 because 3 is less than 5. Describe max error and give the corr

forsale [732]

Just because the 3 is less than 5 doesn't mean it's less. What max did wrong was automatically see the first number and say it was greater/least because it's just the first digit. 36,594,145 is greater than 5,980,251 because the last 6 digits are invisible right now so thing about the numbers before the first comma- 36 and 5. Which number is greater? 36 so just because it's the first digit in the number doesn't make it automatically greater or less. *-sorry for such long paragraph*

7 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.

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