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3 years ago

Find the volume of the rectangular pyramid.

Mathematics

1 answer:

UNO [17]3 years ago

4 0

Answer:

Step-by-step explanation:

V = (1/3)Bh

V = (1/3)(7 ft)(2 ft)(6 ft)

V = 28 ft^3

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Answer Question number 5 <br><br>Plzzzz

Softa [21]

Step-by-step explanation:

Consider x⅓ =a

then the following expression can be written as

a²+a-2

a²+2a-a-2

a(a+2)-1(a+2)

(a+2)(a-1)

a= -2 or, a= 1

Putting the value of a

x⅓ = -2 and, x⅓ =1

6 0

3 years ago

Read 2 more answers

What is the slope of the line in the coordinate plane?

liberstina [14]

Answer:

slope is rise over run

My math teacher says that

5 0

2 years ago

rosy wants a large picture window put in the living room of her new house. The widow is to be square with an area of 49 square f

andre [41]

Each side should be 14.75 square feet long.

4 0

3 years ago

Read 2 more answers

Marcos gana S/18,50 por hora y se le descuenta

Talja [164]

Answer:

$76,9

Step-by-step explanation:

Para calcular cuánto ganó Marcos ese día, tienes que multiplicar el salario por hora por el número de horas que trabajó y restarle el valor por los minutos que llegó tarde que es el resultado de multiplicar los minutos que llegó tarde por el valor a descontar por minuto:

salario: $18,50 por hora

número de horas trabajadas: 5

descuento por tardanza: $1,20 por minuto

número de minutos de tardanza: 13 minutos

salario del día=($18,50*5)-($1,20*13)

salario del día=$92,5-$15,6

salario del día=$76,9

De acuerdo con esto, el salario que Marcos ganó ese día es $76,9.

4 0

3 years ago

(D^2 +2D +1)y=e^-x log(x) solve using method variation of parameter?

Ad libitum [116K]

First you'll need the complementary solutions. The characteristic equation for this ODE is

$D^2+2D+1=(D+1)^2=0\implies D=-1$

with multiplicity 2. This means two linearly independent solutions will be $y_1=e^{-x}$ and $y_2=xe^{-x}$ .

Via variation of parameters, the particular solution will take the form

$y_p=y_1u_1+y_2u_2$

where

$u_1=-\displaystyle\int\frac{y_2e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx$

The Wronskian is

$W(y_1,y_2)=\begin{vmatrix}e^{-x}&xe^{-x}\\-e^{-x}&e^{-x}(1-x)\end{vmatrix}=e^{-2x}(1-x)+xe^{-2x}=e^{-2x}$

So, you have

$u_1=-\displaystyle\int\frac{xe^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx$
$u_1=-\displaystyle\int x\log x\,\mathrm dx$
$u_1=\dfrac14x^2-\dfrac12x^2\log x$

and

$u_2=\displaystyle\int\frac{e^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx$
$u_2=\displaystyle\int\log x\,\mathrm dx$
$u_2=x(\log x-1)$

So the solution to the ODE is

$y=C_1y_1+C_2y_2+y_1u_1+y_2u_2$
$y=(C_1+C_2x)e^{-x}+\left(\dfrac14x^2-\dfrac12x^2\log x\right)e^{-x}+(\log x-1)x^2e^{-x}$
$y=(C_1+C_2x)e^{-x}+\dfrac14x^2e^{-x}(2\log x-3)$

3 0

3 years ago