Mathematics: I need help with this geometry question

I need help with this geometry question - id10833540, christianfielding336

Ответ:

cooboy

18.03.2022

Mathematics

x=9; JL= 23

Step-by-step explanation:

To find x:

If JL and KM are equivalent, then JL = KM. Substitute the equations and you get 2x+5 = 7x-40.

Put the similar terms on the same side by subtracting 2x from both sides and adding 40 to both sides. You will get 45=5x.

Get x alone by dividing both sides by 5 and you will get 9=x.

To find JL:

Substitute the x in 2x+5 (JL length equation) with 9. Your new equation will be 2(9) + 5.

2•9 = 18.

18+5=23

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Ответ:

donterriuscollier

19.03.2021

Mathematics

g(x) = 4x² - x

General Formulas and Concepts:

Algebra II

Functions

Function NotationPiecewise Functions

Calculus

Integration

IntegralsIntegral NotationIntegration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Step-by-step explanation:

*Note:

Integrating a piecewise function requires you to integrate both parts.

Step 1: Define

Identify.

$\displaystyle f(x) = \left \{ {{8x - 1 ,\ x \leq 4} \atop {31 ,\ x \geq 4}} \right.$

$\displaystyle \int {f(x)} \, dx = \left \{ {{g(x) + C ,\ x \leq 4} \atop {31x + C ,\ x \geq 4}} \right.$

Step 2: Find function g(x)

We can see that the 2nd part of the piecewise function already has been integrated:

[Integral] Set up: $\displaystyle \int {f(x)} \, dx ,\ x \geq 4 = \int {31} \, dx ,\ x \geq 4$ [Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {f(x)} \, dx ,\ x \geq 4 = 31 \int {} \, dx ,\ x \geq 4$ [Integral] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle \int {f(x)} \, dx ,\ x \geq 4 = 31x + C ,\ x \geq 4$

To find function g(x), we simply have the same setup:

[Integral] Set up: $\displaystyle \int {f(x)} \, dx ,\ x \leq 4 = \int {8x - 1} \, dx ,\ x \leq 4$ [Integral] Rewrite [Integration Rule - Addition/Subtraction]: $\displaystyle \int {f(x)} \, dx ,\ x \leq 4 = \int {8x} \, dx - \int {1} \, dx ,\ x \leq 4$ [Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {f(x)} \, dx ,\ x \leq 4 = 8 \int {x} \, dx - \int {} \, dx ,\ x \leq 4$ [Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle \int {f(x)} \, dx ,\ x \leq 4 = 8 \bigg( \frac{x^2}{2} \bigg) - x + C ,\ x \leq 4$ Simplify: $\displaystyle \int {f(x)} \, dx ,\ x \leq 4 = 4x^2 - x + C ,\ x \leq 4$ Redefine: $\displaystyle g(x) = 4x^2 - x + C ,\ x \leq 4$

The integration constant C is already included in the answer, so our answer is g(x) = 4x² - x.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration (Applications)

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