AB=A, B, equals Round your answer to the nearest hundredth.

The length of a rectangle is 5 ft less than three times the width, and the area of the rectangle is 50 ft². Find the dimensions

Mariulka [41]

Answer

Length = 10 ft

Width = 5 ft

Explanation

Area of the rectangle given = 50 ft²

Let the width of the rectangle be x

So this means the length of the rectangle will be 3x - 5

What to find:

The dimensions of the rectangle.

Step-by-step solution:

Area of a rectangle = length x width

i.e A = L x W

Put A = 50, L = 3x - 5, W = x into the formula.

$\begin{gathered} 50=(3x-5)x \\ 50=3x^2-5x \\ 3x^2-5x-50=0 \end{gathered}$

The quadratic equation can now be solve using factorization method:

$\begin{gathered} 3x^2-5x-50=0 \\ 3x^2-15x+10x-50=0 \\ 3x(x-5)+10(x-5)=0 \\ (3x+10)(x-5)=0 \\ 3x+10=0\text{ }or\text{ }x-5=0 \\ 3x=-10\text{ }or\text{ }x=5 \\ x=-\frac{10}{3}\text{ }or\text{ }x=5 \end{gathered}$

Since the dimension can not be negative, hence the value of x will be = 5.

Therefore, the dimensions of the rectangle will be:

$\begin{gathered} Length=3x-5=3(5)-5=15-5=10\text{ }ft \\ \\ Width=x=5\text{ }ft \end{gathered}$

7 0

1 year ago

Algebra 2 Please Help

hammer [34]

No sé lo que quieres decir

5 0

3 years ago

Hamid's soccer game will start at 11:00 A.M., but the players must arrive at the field three-quarters of an hour early to warm u

Minchanka [31]

If the game will start at 11:00 A.M., but the players must arrive at the field three-quarters of an hour early to warm up, it refers to 8:45 a.m. Why? If we start to count in 11 backward and start to trace the three-quarters, it shows that 10:45, 9:45, and 8:45 are the three-quarters. So Hamid statement that he has to be at the field at 9:45 A.M is not correct.

4 0

3 years ago

Point B has coordinates (4,1). The x-coordinate of point A is -4. The distance between point A and point B is 10 units. What a

klasskru [66]

Given:

Point B has coordinates (4,1).

The x-coordinate of point A is -4.

The distance between point A and point B is 10 units.

To find:

The possible coordinates of point A.

Solution:

Let the y-coordinate of point A be y. Then the two points are A(-4,y) and B(4,1).

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The distance between point A and point B is 10 units.

$\sqrt{(4-(-4))^2+(1-y)^2}=10$

Taking square on both sides, we get

$(8)^2+(1-y)^2=100$

$(1-y)^2=100-64$

$(1-y)^2=36$

Taking square root on both sides, we get

$(1-y)=\pm \sqrt{36}$

$-y=\pm 6-1$

$y=1\mp 6$

$y=1-6$ and $y=1+6$

$y=-5$ and $y=7$

Therefore, the possible coordinates of point A are either (-4,-5) or (-4,7).

7 0

2 years ago

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

7 0

3 years ago