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2 years ago

¿Cuál de los siguientes productos algebraicos es un Binomio Suma por Binomio Diferencia?

Mathematics

1 answer:

Butoxors [25]2 years ago

3 0

The answer is d hope that helps

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What is the surface area of this triangular pyramid?

solniwko [45]

Answer:

Step-by-step explanation:

Hence, the surface area of a triangular pyramid is 140 square units.

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2 years ago

Renee bought a magazine subscription for a year. She paid $27. Renee wanted to find the price, p, of the subscription each month

kirza4 [7]

The price is 2.25, I found that answer by dividing 27 and 12 for that it would be 12p=27 divide 12 from both sides, p=2.25

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2 years ago

The area of a square is 9 centimeters squared (cm^2) If the sides of the square are doubled in length, what is the new area, in

Mice21 [21]

Answer:

The question says area of a square = 9 cm squared.

And they said if the sides are doubled in length what is the new length

so new length = 9 cm^2 × 9 cm^2

which should give you 81cm^2

therefore your answer should be = C

7 0

2 years ago

Read 2 more answers

1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago

Find the remainder when

Alona [7]

Dividing f(x) by 2x + 5 leaves the same remainder as division by x + 5/2. By the remainder theorem, it is equal to f (-5/2), so the remainder here is

f (-5/2) = 8 (-5/2)³ + 4 (-5/2)² - 13 (-5/2) + 3 = -129/2

6 0

3 years ago