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3 years ago

What is the surface are of the triangle prism

Mathematics

1 answer:

stepladder [879]3 years ago

5 0

You can find the area of a triangular prism by putting it into a bet and finding the area of all sides then adding them together

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Find the value of x in the equation x⁄ 4 + x⁄14 + x⁄17 = 71

Morgarella [4.7K]

Hello from MrBillDoesMath!

Answer:

33769/181

Discussion:

Each term contains "x" so factoring it out gives

x( 1/4 + 1/14 + 1/17) = 71 (*)

Use common factor (17*14*4 = 952) as the denominator to combine terms:

1/4 = (17*14)/ 952 = 238/952

1/14 = (17*4)/952 = 68/952

1/17 = (14*4)/952 = 56/952

so 1/4 + 1/14 + 1/17 = (238 + 68 + 56)/ 952 = 362/952 = 181/476

Substituting in (*) gives

x ( 181/476) = 71 => multiply both sides by 476/181

x = (71 * 476)/181 => 71* 476 =33769

x = 33769/181

Thank you,

MrB

4 0

3 years ago

Is 2 an irrational number

Lena [83]

Answer:

no its rational

Step-by-step explanation:

3 0

3 years ago

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How do I solve 4(5(6))<br><br> This is a hypothetical question

SSSSS [86.1K]

PEMDAS states that the value in the parenthesis should be evaluated first:
4(5*6)
4(30)
After just multiply the value in the parenthesis by the number outside of the parenthesis:
120

6 0

3 years ago

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(5⅐)⁴=( ) OPTIONS : A.) (5⁴)⅐ B.)5²⁸ C.)(5)¹/²⁸ D.)(5)⁷/⁴

NemiM [27]

Answer:

Step-by-step explanation:

4*1/7

5 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

3 years ago