Ask question

Ask question

All categories

3 years ago

7

the area of a rectangular room is 750 square feet. the width of the room is 5 feet less than the length of the room. which equat

ions can be used to solve for y, the length of the room? check all that apply. y(y 5) = 750 y2 – 5y = 750 750 – y(y – 5) = 0 y(y – 5) 750 = 0 (y 25)(y – 30) = 0

2 answers:

pogonyaev3 years ago

8 0

the correct answer is

B)y2 – 5y = 750

C)750 – y(y – 5) = 0

E)(y + 25)(y – 30) = 0

I JUST took the test

zhuklara [117]3 years ago

3 0

Let's assume length of the room as y
so the width of the room = y - 5

Area
y * (y - 5) = 750

You might be interested in

When working in trigonometry, you typically measure angles in

Neporo4naja [7]

Answer:

Radians

<em>Hope that helps!</em>

Step-by-step explanation:

8 0

3 years ago

What is the value of this expression if h<br> = 8,j = -1, and k<br> =<br> -12

Rashid [163]

Answer:

12

Step-by-step explanation:

The given expression is :

$\dfrac{j^3k}{h^0}$

We need to find the value of this expression when h = 8, j = -1 and k = -12.

Put all the values in the given expression.

$\dfrac{j^3k}{h^0}\\\\=\dfrac{(-1)^3\times (-12)}{(8)^0}\\\\=\dfrac{-1\times -12}{1}\\\\=12$

So, the value of the given expression is 12.

7 0

3 years ago

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)

ELEN [110]

Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

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

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

3 0

3 years ago

How do I write an expression for n . a using only addition

Taya2010 [7]

Answer:

i don't know anymore

Step-by-step explanation: I don't know anymore

3 0

2 years ago

Change 3/8 in to decimal show working

mixas84 [53]

Answer:

0.375

Step-by-step explanation:

Convert the fraction to a decimal by dividing the numerator by the denominator.

3/8 = 0.375

7 0

3 years ago