Using the following triangle, what is the cosine of angle a?

Two kinds of tickets to an outdoor concert were sold: lawn tickets and seat tickets. Less than 400 tickets in total were sold. L

miskamm [114]

Answer:

Tickets to sit on the bench (b) = 150

Tickets to sit on the lawn (l) = 200

Step-by-step explanation:

Tickets to sit on a bench (b)cost $75 each.

Tickets to sit on the lawn (l) cost $40 each

350 tickets had been sold

$19,250 had been raised through tickets sales

This forms a simultaneous equation:

b + l = 350 ... (i)

75b + 40l = 19,250 ... (ii)

Multiplying (i) by 40 and (ii) by 1 we get;

40b + 40l = 14,000 ... (i)

75b + 40l = 19,250 ... (ii)

Subtracting (ii) - (i) we get;

35b = 5250

b = 5250 ÷ 35 = 150

7 0

3 years ago

State the value of the discriminant for <br><br>y = <img src="https://tex.z-dn.net/?f=5x%5E2%2B9x-3" id="TexFormula1" title="5x^

Nostrana [21]

Answer:

141, meaning that there will be two real solutions

Step-by-step explanation:

The discriminant of a quadratic is $b^2-4ac$ , which in this case is:

$9^2-4(5)(-3)=81+60=141$

Since the discriminant is positive and not zero, there will be two real solutions to this equation. This is because when the discriminant is negative, and you take the square root of it, you get a negative number. If you take the square root of 0, you get 0, which means that there will only be one solution to the equation.

Hope this helps!

3 0

3 years ago

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

5 0

3 years ago

Please help. I’ll mark you as brainliest if correct!

sesenic [268]

Answer:

Quantity (lbs) of type 1 candy x = 8

Quantity (lbs) of type 2 candy y = 17,5

Step-by-step explanation:

Let´s call "x" quantity (in pounds) of candy type 1 in the mixture, and "y" quantity (in pounds ) of candy type 2, then according to the problem statement.

x + y = 25,5

2,20*x + 7,30*y = 5,70 * 25,5 ⇒ 2,20*x + 7,30*y = 145,35

Then we have a two equation system

x + y = 25,5 ⇒ y = 25,5 - x

2,20*x + 7,30*y = 145,35 ⇒ 2,20*x + 7,30* (25,5 - x ) = 145,35

2,20*x + 186,15 - 7,30*x = 145,35

5,1*x = 40,8

x = 40,8/5,1

x = 8 lbs

And y = 25,5 - 8

y = 17,5 lbs

5 0

3 years ago

How many prime numbers are there between 45 and 72?

laila [671]

There are 47,51,53,59,61,67,71, that is, 7 primes.

4 0

3 years ago

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