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4 years ago

15

Solve for side in right triangles

1 answer:

IgorLugansk [536]4 years ago

8 0

Answer:

tbh idk im not so good at math

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What is the length of x

GrogVix [38]

Answer:

The answer is 1 3/5

Step-by-step explanation:

2÷5 = 0.4
0.4×4 = 1.6
1.6 = 1 3/5

8 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

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4 0

2 years ago

Solving Quadratic Equations

NISA [10]

Answer:

a) x= 5 or x= -3

b) x = -4 + $\sqrt{26}$ or -4 - $\sqrt{26}$

Step-by-step explanation:

a) $x^{2}$ - 2x - 15 =0

x = $\frac{-(-2) +\sqrt{(-2)^{2} - 4(1)(-15)}}{2(1)}$ or $\frac{-(-2) -\sqrt{(-2)^{2} - 4(1)(-15)}}{2(1)}$

x= 5 or x=-3

b) $x^{2}$ + 8x - 10 =0

x = $\frac{-(8) +\sqrt{(8)^{2} - 4(1)(-10)}}{2(1)}$ or $\frac{-(8) -\sqrt{(8)^{2} - 4(1)(-10)}}{2(1)}$

x = -4 + $\sqrt{26}$ or -4 - $\sqrt{26}$

7 0

3 years ago

The correct ordered pair 2y + 4x equals 18 2y - 3x equals 4

Julli [10]

Answer:

I don’t get it? What’s the question?

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers

Determine the dimensions of a rectangular solid (with a square base) with maximum volume if its surface area is 181.5 square cen

3241004551 [841]

Answer:

The base and height of the solid is 5.5cm

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Step-by-step explanation:

Given

$Surface\ Area = 181.5cm^2$

Required

Determine the dimensions that maximizes the volume

Let the base dimension be x and the height be h

The volume is calculated as:

$Volume =x^2 * h$

$Volume =x^2h$

181.5 =x^2h

The surface area (S) is calculated as this:

$S = 2(x^2 + xh + xh)$

$S = 2(x^2 + 2xh)$

$S = 2x^2 + 4xh$

Substitute 181.5 for S

$181.5 = 2x^2 + 4xh$

Make h the subject:

$4xh = 181.5 - 2x^2$

$h = \frac{181.5 - 2x^2}{4x}$

Substitute $h = \frac{181.5 - 2x^2}{4x}$ in $Volume =x^2h$

$V = x^2(\frac{181.5 - 2x^2}{4x})$

$V = x(\frac{181.5 - 2x^2}{4})$

$V = \frac{1}{4}(x)(181.5 - 2x^2)$

$V = \frac{181.5x}{4} - \frac{2x^3}{4}$

$V = \frac{181.5x}{4} - \frac{x^3}{2}$

To get the maximum, we differentiate V with respect to t and set the differentiation to 0

$dV = \frac{181.5}{4} - \frac{3x^2}{2}$

Set to 0

$0 = \frac{181.5}{4} - \frac{3x^2}{2}$

$\frac{3x^2}{2} = \frac{181.5}{4}$

Multiply through by 4

$4 * \frac{3x^2}{2} = \frac{181.5}{4} * 4$

$2*3x^2 = 181.5$

$6x^2 = 181.5$

$x^2 = \frac{181.5}{6}$

$x^2 = 30.25$

$x = \sqrt{30.25$

$x = 5.5$

Recall that:

$h = \frac{181.5 - 2x^2}{4x}$

$h = \frac{181.5 - 2 * 5.5^2}{4 * 5.5}$

$h = \frac{181.5 - 60.5}{22}$

$h = \frac{121}{22}$

$h = 5.5$

So, we have:

$h = 5.5$

$x = 5.5$

<em>Hence, the base and height of the solid is 5.5cm</em>

7 0

3 years ago