How to complete the problem

Andrews [41]

The piecewise defined function is formed by the following three functions: y = 40, for 0 ≤ x < 15, y = 0.0425 · x² - 3.8 · x + 87.438, for 15 ≤ x < 65 and y = 0.4 · x - 6, for x ≥ 65.

<h3>What is the piecewise defined function behind the graph?</h3>

In this problem we have a piecewise defined function formed by three functions: (i) Linear equation, (ii) Quadratic equation, (iii) Linear equation. The first function is y = 40, for 0 ≤ x < 15.

The second function can be found based on the knowledge of three points:

y = a · x² + b · x + c

(x₁, y₁) = (15, 40)

40 = 225 · a + 15 · b + c

(x₂, y₂) = (45, 2.5)

2.5 = 2025 · a + 45 · b + c

(x₃, y₃) = (65, 20)

20 = 4225 · a + 65 · b + c

And the quadratic equation is y = 0.0425 · x² - 3.8 · x + 87.438, for 15 ≤ x < 65.

And the latter linear equation is:

y = m · x + b

m = (30 - 20)/(90 - 65)

m = 0.4

And the x-intercept is:

b = y - m · x

b = 20 - 0.4 · 65

b = - 6

The linear equation is y = 0.4 · x - 6, for x ≥ 65.

The piecewise defined function is formed by the following three functions: y = 40, for 0 ≤ x < 15, y = 0.0425 · x² - 3.8 · x + 87.438, for 15 ≤ x < 65 and y = 0.4 · x - 6, for x ≥ 65.

To learn more on piecewise defined functions: brainly.com/question/12561612

#SPJ1

7 0

2 years ago

Question 12 With steps please

34kurt

$Answer:\ \boxed{5^2}$

$\sqrt[n]{a}=a^\frac{1}{n}\\\\\text{therefore}\\\\\sqrt5=5^\frac{1}{2}\\----------------\\25=5^2\\----------------\\78,125=5^7\\----------------\\\text{use}\\\\a^n\cdot a^m=a^{n+m}\\\\\dfrac{a^n}{a^m}=a^{n-m}\\----------------\\\\\dfrac{\sqrt5\times78,125}{25\times5^{\frac{7}{2}}}=\dfrac{5^\frac{1}{2}\times5^7}{5^2\times5^\frac{7}{2}}=\dfrac{5^{\frac{1}{2}+7}}{5^{2+3\frac{1}{2}}}=\dfrac{5^{7\frac{1}{2}}}{5^{5\frac{1}{2}}}=5^{7\frac{1}{2}-5\frac{1}{2}}=\boxed{5^2}$

6 0

3 years ago

What is the quotient 2y^-6y-20/4y+12 ÷ y+5y+6/3y^2+28y+27

malfutka [58]

Answer with explanation:

$\rightarrow \frac{\frac{2y^2-6 y-20}{4 y+12}}{\frac{y^2+5 y+6}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{y^2-3y-10}{2 y+6}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{(y-5)(y+2)}{2 (y+3)}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{(y-5)(y+2)}{2 (y+3)}} \times {\frac{3 y^2+28 y+27}{(y+2)(y+3)}}\\\\ \rightarrow\frac{(y-5)\times(3 y^2+28 y+27)}{2 (y+3)^2}}$

→y²+5y+6

=y²+3 y+2 y+6

=y×(y+3)+2×(y+3)

=(y+2)(y+3)

→y² -3 y-10

=y² -5 y+2 y -10

=y×(y-5)+2×(y-5)

=(y+2)(y-5)

6 0

3 years ago

Read 2 more answers

The length of one side of a square field is represented by the expression 3-7x.

Juliette [100K]

They give us the length of side of the square which is 3-7x.

To get an expression for the perimeter that is expressed as a product we have to multiply the expression 3-7x by four since every square has four sides. This will give us an expression of 4(3-7x). We than can use the distributive property to multiply everything inside the parenthesis by 4. This will give us an expression of 12-28x.

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=5x%5E%7B2%7D%20%2B%20x-7%3D" id="TexFormula1" title="5x^{2} + x-7=" alt="5x^{2} + x-7=" align=

zysi [14]

Equation: $5x^2+x-7$

Since there aren't any like terms and it can't be factored, your answer would be:

$5x^2+x-7$

Hope this answer helped you!

7 0

3 years ago