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3 years ago

1.1Solve the following equations:1.1.1 (x+5)(x-3)=33

Mathematics

1 answer:

schepotkina [342]3 years ago

8 0

Step-by-step explanation:

x²-3x +5x -15=33

x²-3x+5x=33+15

x²-3x+5x=48

x²+2x=48

divide both sides by 2

x²+x =24

x²= 24

x= ✓24

dunno from here

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Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago

PART1

Mice21 [21]

Part 1: $\frac{5x^{3} + 10x^{2} + 15x}{5x} = \frac{5x^{3}}{5x} + \frac{10x^{2}}{5x} + \frac{15x}{5x} = x^{2} + 2x + 3$
The answer is C.

Part 2: $\frac{3x^{4} + 5x^{2} + 2x}{x} = \frac{3x^{4}}{x} + \frac{5x^{2}}{x} + \frac{2x}{x} = 3x^{3} + 5x + 2$
The answer is A.

8 0

4 years ago

Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2

iren2701 [21]

Hello there, hope I can help!

I assume you mean $L\left\{\frac{ekt+e-kt}{2}\right\}$
With that, let's begin

$\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \ L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}$

$\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}$
$\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b$
$L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}$
$\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}$

$L\left\{t\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \ L\left\{t\right\}=\frac{1}{s^2}$

$L\left\{\frac{e}{2}\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \ L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \ \frac{e}{2s}$

$\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}$

$\frac{ek}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{ek\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{ek}{2s^2}$

$\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{k\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{k}{2s^2}$

$\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}$

Hope this helps!

3 0

4 years ago

We know that a triangle with side lengths , , and is a right triangle. Using those side lengths, find the missing triples and x-

Anit [1.1K]

The completed table for the Pythagorean triples are:

$x \text{ value } \text{ Pythagorean triple}\\\text{ } \text{ } \text{ } 3 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (6,8,10)\\\text{ } \text{ } \text{ } 4 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (8,15,17)$

$\text{ } \text{ } \text{ } 5 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (10,24,26)\\\text{ } \text{ } \text{ } 6 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (12,35,37)$ .

A trio of positive numbers known as a Pythagorean triple fits into the Pythagoras theorem's formula, which is written as a² + b² = c², where a, b, and c are all positive integers. Here, "a" and "b" make up the other two legs of the right-angled triangle, and "c" serves as the triangle's "hypotenuse," or longest side. In terms of the Pythagorean triples (a, b, c).

In the question, we are given that a triangle with side lengths x² - 1, 2x, x² + 1 is a right triangle, and are asked to fill in the missing table:

$x \text{ value } \text{ Pythagorean triple}\\\text{ } \text{ } \text{ } 3 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \underline{\text{ }}\\\text{ } \text{ } \text{ } \underline{\text{ }} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (8,15,17)\\$

$\text{ } \text{ } \text{ } 5 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \underline{\text{ }}\\\text{ } \text{ } \text{ } \underline{\text{ }} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (12,35,37)$ .

Taking x as 3, and putting in the sides, we get:

x² + 1 = 3² + 1 = 10.

2x = 2*3 = 6.

x² - 1 = 3² - 1 = 8.

Thus, the triplet is 6,8,10.

For the triplet 8,15,17, we get the x value as:

2x = 8,

or, x = 8/2 = 4.

Thus, the x-value for the triplet (8,15,17) is 4.

Taking x as 5, and putting in the sides, we get:

x² + 1 = 5² + 1 = 26.

2x = 2*5 = 10.

x² - 1 = 5² - 1 = 24.

Thus, the triplet is 10,24,26.

For the triplet 12,35,37, we get the x value as:

2x = 12,

or, x = 12/2 = 6.

Thus, the x-value for the triplet (12,35,37) is 6.

Thus, the completed table for the Pythagorean triples are:

Learn more about the Pythagorean triplets at

brainly.com/question/24211694

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The provided question is incomplete. The complete question is:

"We know that a triangle with side lengths x² - 1,2x and x² + 1 is a right triangle. Using those side lengths, find the missing triples and x-values.

Write the triples in parentheses, without spaces between the numbers, and with a comma between numbers. Write the triples in order from least to greatest.

Type the correct answer in each box.

$x \text{ value } \text{ Pythagorean triple}\\\text{ } \text{ } \text{ } 3 \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \underline{\text{ }}\\\text{ } \text{ } \text{ } \underline{\text{ }} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } (8,15,17)\\$

6 0

2 years ago

How many inches are in 35 feet?

soldier1979 [14.2K]

Answer:

420 inches

Step-by-step explanation:

There are twleve inches in one foot.

(35) (12) = 420

4 0

4 years ago

Read 2 more answers

1.1Solve the following equations:1.1.1 (x+5)(x-3)=33​

1.1Solve the following equations:1.1.1 (x+5)(x-3)=33