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3 years ago

A) Find the value of x.

Mathematics

1 answer:

Tcecarenko [31]3 years ago

5 0

It is a pentagon. The sum of all angles of pentagon is 540°.

A) $(x+10)+(2x-10)+(x-25)+(x+20)+(x-13)=540$
$x=93$

B) x+20 = 93+20 = 113°. The adjacent exterior angle is 180 - 113 = 67°

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Plzplz will make bianleast Find the volume in cubic meters.

galben [10]

Answer:

0.000005832 m³

Step-by-step explanation:

V = L²

= (18mm)³

= 5832 mm³

= 0.000005832 m³

8 0

3 years ago

Read 2 more answers

What is the factored form of the expression?<br><br> d^2 = 36d + 324

Veseljchak [2.6K]

Answer:

$(x-43.45)(x+7.45)=0$

Step-by-step explanation:

We have the quadratic equation $d^{2}=36d+324$ i.e. $d^{2}-36d-324=0$

As, the roots of the quadratic equation $ax^{2}+bx+c=0$ are given by $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ .

So, from the given equation, we have,

a = 1, b = -36 , c = -324.

Substituting the values in $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ , we get,

$x=\frac{36\pm \sqrt{(-36)^{2}-4\times 1\times (-324)}}{2\times 1}$

i.e. $x=\frac{36\pm \sqrt{1296+1296}}{2}$

i.e. $x=\frac{36\pm \sqrt{2592}}{2}$

i.e. $x=\frac{36\pm 50.9}{2}$

i.e. $x=\frac{36+50.9}{2}$ and $x=\frac{36-50.9}{2}$

i.e. $x=\frac{86.9}{2}$ and $x=\frac{-14.9}{2}$

i.e. x = 43.45 and x = -7.45

Thus, the roots of the equation are 43.45 and -7.45.

Hence, the factored form of the given expression will be $(x-43.45)(x+7.45)=0$

3 0

3 years ago

(2x³+4x²-x+8) / (x² + 3x + 2)

a_sh-v [17]

$\dfrac{2x^3+4x^2 -x+8}{x^2 +3x+2}\\\\\\=\dfrac{2x^3+6x^2+4x-2x^2-6x-4+x+12}{}\\\\=\dfrac{2x(x^2 +3x+2)-2(x^2 +3x+2)+x+12}{x^2 +3x+2}\\\\\\=\dfrac{2x(x^2+3x+2)}{x^2 +3x +2 }- \dfrac{2(x^2 +3x +2)}{x^2 +3x +2 } + \dfrac{x+12}{x^2 +3x+2}\\\\\\=2x-2 + \dfrac{x+2}{x^2 +3x +2}\\\\\\=2x-2 + \dfrac{x+12}{x^2 +2x +x +2}\\\\\\=2x-2 + \dfrac{x+12}{x(x+2) + x+2}\\\\\\=2x-2 + \dfrac{x+12}{(x+1)(x+2)}\\$

3 0

2 years ago

How many times does 12 go into 60?

statuscvo [17]

Answer:

5 times

Step-by-step explanation:

3 0

3 years ago

The gas co2 is diffusin at steady state through a tube 20 cm long having a diameter of 1 cm and containing n2 at 350 k . the tot

Marrrta [24]

The value is J = 1.71*10^-6 kmol/m^2.s

According to the given conditions we get,

The length the tube l= 0.20m

The diameter of the tube d= 0.01m

The total pressure inside the tube P= 101.32kPa

The partial pressure of CO2 at the first end is

P1= 456mm Hg = 60794.832 Pa

The partial pressure of CO2 at the other end is

P2= 76mm Hg = 10132.472 Pa

The temperature is T = 298 K

The diffusion coefficient D= 1.67* 10^-5 m^2/s

Generally the molar flux of CO2 is mathematically represented as

J= D{p1-p2} / R.T

Here R is the gas constant with value R= 8.314 j/kmol

So we get J= 1.71 * 10^-3 mol/m^2.sec

Learn more about Mole on:

brainly.com/question/27270616

#SPJ4

4 0

2 years ago