18g + 24 write the expression in two other ways.

A pool measuring 18 meters by 22 meters is surrounded by a path of uniform width. It means that the combined length of the pool and the uniform path is (18 + 2x) meters and the combined width of the pool and the uniform path is (22 + 2x) meters.

If the area of the pool and the path combined is 1440 square meters, it means that

(18 + 2x)(22 + 2x) = 1440

396 + 36x + 44x + 4x² = 1440

4x² + 80x + 396 - 1440 = 0

4x² + 80x - 1044 = 0

Dividing both sides of the equation by 4, it becomes

x² + 20x - 261 = 0

x² + 29x - 9x - 261 = 0

x(x + 29) - 9(x + 29) = 0

x - 9 = 0 or x + 29 = 0

x = 9 or x = 29

Since the width cannot be negative, then x = 9 meters

5 0

3 years ago

You have dimes and quarters in your pocket. There are 12 coins that total $2.25. Write and solve a system of linear equations to

icang [17]

Answer:

Hence the System of equation are $\left \{ {{x+y=12} \atop {0.1x+0.25y=2.25}} \right.$

There are 5 dimes and 7 quarters in my pocket.

Step-by-step explanation:

Let Number of dimes be 'x'.

Also Number of quarters be 'y'.

Now Given:

Total Number of Coins = 12

So the equation can be framed as;

$x+y=12 \ \ \ \ equation \ 1$

Also Given:

Total Amount in pocket = $2.25

Now we know that 1 dime = $0.1

Also 1 quarter =$0.25

So the equation can be framed as;

$0.1x+0.25y = 2.25 \ \ \ \ equation \ 2$

Hence the System of equation are $\left \{ {{x+y=12} \atop {0.1x+0.25y=2.25}} \right.$

Now Solving the equation we get;

Now Multiplying equation 2 by 10 we get;

$10(0.1x+0.25y)=2.25\times10\\\\10\times0.1x+10\times0.25y=22.5\\\\x+2.5y=22.5 \ \ \ \ equation\ 3$

Now Subtracting equation 1 from equation 3 we get;

$(x+2.5y)-(x+y)=22.5-12\\\\x+2.5y-x-y =10.5\\\\1.5y =10.5\\\\y= \frac{10.5}{1.5}= 7$

Now Substituting the value of y in equation 1 we get;

$x+y=12\\\\x+7=12\\\\x=12-7 =5$

Hence there are 5 dimes and 7 quarters in my pocket.

6 0

3 years ago

Let , 4 ,− 7 be a point on the terminal side of θ . find the exact values of cos θ , csc θ , and tan θ .

Makovka662 [10]

Ok, you are given a point and you need to find the exact values for $cos \theta csc \theta & tan \theta$

First thing first. We need to see if we are working with a unit circle and find the radius.

How to tell if we are working with a unit circle?
We know $x^2 + y^2 = r^2$ is a circle.

We know that to find the radius we can use the following formula:
$r^2 = \sqrt{x^2 + y^2}$

If $r^2 = \sqrt{x^2 + y^2} = 1$ we are working with a unit circle.

Lets see if it = 1.
$r^2 = \sqrt{4^2 + -7^2}$
$r^2 = \sqrt{16 + 49}$
$r^2 = \sqrt{65}$

Square both sides now
$\sqrt{r^2} = \sqrt{\sqrt{65}}$
$r = \pm 65^{\frac{1}{4}}}$

Since we squared, we have a + and a - but we disregard the - because we do not have - radii
$r = 65^{\frac{1}{4}}}$
We can also say
$r = \sqrt[4]{65}$

Ok, since r does not equal 1, we are not working with a unit circle but we have found r, which is our radius.

Now that we know the value of r, which is $r = 65^{\frac{1}{4}}}$ , we need to look at the identities of cos, csc and tan.

The identities:
$cos \theta = \frac{x}{r}$
$csc \theta = \frac{1}{y}$
$tan \theta = \frac{y}{x}$

Now that we know their identities and know the radius of our circle, we can find the exact values of cos, csc and tan.

$cos \theta = \frac{x}{r} = \frac{4}{65^{\frac{1}{4}}}$
$csc \theta = \frac{1}{y} = \frac{1}{-7}$
$tan \theta = \frac{y}{x} = \frac{-7}{4}$

The exact values for cos, csc and tan given the point (4,-7) are:
$cos \theta = \frac{4}{65^{\frac{1}{4}}}$
$csc \theta = \frac{1}{-7}$
$tan \theta = \frac{-7}{4}$

4 0

3 years ago