Subtract 3 from both sides so that the equation becomes -2x^2 + 5x - 13 = 0.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -5 ± √((5)^2 - 4(-2)(-13)) ] / ( 2(-2) )
x = [-5 ± √(25 - (104) ) ] / ( -4 )
x = [-5 ± √(-79) ] / ( -4)
Since √-79 is nonreal, the answer to this question is that there are no real solutions.
Answer:
Step-by-step explanation:
Given the definite integral 
, we to evaluate it. Using integration by substitution method.
Let u = 1-2x⁵ ...1
du/dx = -10x⁴
dx = du/-10x⁴.... 2
Substitute equation 1 and 2 into the integral function and evaluate the resulting integral as shown;
![= \dfrac{-1}{10} \int\limits {\dfrac{du}{u^5} } \\\\= \dfrac{-1}{10} \int\limits {{u^{-5}du } \\= \dfrac{-1}{10} [{\frac{u^{-5+1}}{-5+1}] \\\\= \dfrac{-1}{10} ({\frac{u^{-4}}{-4})\\\\](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B-1%7D%7B10%7D%20%5Cint%5Climits%20%7B%5Cdfrac%7Bdu%7D%7Bu%5E5%7D%20%7D%20%20%5C%5C%5C%5C%3D%20%5Cdfrac%7B-1%7D%7B10%7D%20%5Cint%5Climits%20%7B%7Bu%5E%7B-5%7Ddu%20%7D%20%20%5C%5C%3D%20%5Cdfrac%7B-1%7D%7B10%7D%20%5B%7B%5Cfrac%7Bu%5E%7B-5%2B1%7D%7D%7B-5%2B1%7D%5D%20%20%5C%5C%5C%5C%3D%20%5Cdfrac%7B-1%7D%7B10%7D%20%28%7B%5Cfrac%7Bu%5E%7B-4%7D%7D%7B-4%7D%29%5C%5C%5C%5C)
substitute u = 1-2x⁵ into the result
Hence
Answer:
<h3>
A = ²⁵/₄x² + ⁷⁵/₂x + 50</h3>
Step-by-step explanation:
L = ⁵/₂x + 10
W = ⁵/₂x + 5
A = L•W
A = (⁵/₂x + 10)(⁵/₂x + 5)
A = ⁵/₂x•⁵/₂x + ⁵/₂x•5 + 10•⁵/₂x + 10•5
A = ²⁵/₄x² + ²⁵/₂x + ⁵⁰/₂x + 50
A = ²⁵/₄x² + ⁷⁵/₂x + 50
Or if yoy mean:
L = 5/(2x) + 10
W = 5/(2x) + 5
A = [5/(2x) + 10][5/(2x) + 5] = 25/(4x²) + 75/(2x) + 50
Answer:
Checked on edg. , the answer is A. Thanks!
Step-by-step explanation:
Let x = total sales of baked goods remember 20% = .20 as a decimal
so the equation would be .20x = $600
divide both sides by .20
.20x/.20 = $600/.20
x = $3000 the total sales of baked goods
CHECK: 20% times $3000 = .20 times $3000 = $600
