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3 years ago

Colin drew the floor plan of his bedroom shown below. What is the area of his bedroom in square feet?

Mathematics

1 answer:

labwork [276]3 years ago

4 0

512 because multiply each number together

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Sherry is a waitress at a country club restaurant. She receives an automatic 15% tip which is added to the member's check for th

Karo-lina-s [1.5K]

Answer:

143 50

Step-by-step explanation:

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2 years ago

Lesson 4 Solving One Variable Equation

aleksandr82 [10.1K]

Answer:

b = -³¹⁄₃

Step-by-step explanation:

We can solve this Algebra equation by separating the variable and constants.

3b + 15 = -26 - 20

3b + 15 = -46

3b = -31

b = -³¹⁄₃

3 0

3 years ago

Y=-2x²-8x-5 domain and range

Jobisdone [24]

Answer:

Domain: All real numbers

Range: y≤3

Step-by-step explanation:

The given function is

$y = - 2 {x}^{2} - 8x - 5$

This is a polynomial function.

The domain is set of all values of x, that makes this function defined.

Since polynomial functions are defined everywhere, the domain is all real numbers.

To find the range we put the function in vertex form:

$y = - 2( {x}^{2} + 4x) - 5$

$y = - 2( {x}^{2} + 4x + 4) - 5 + - 4 \times - 2$

$y = - 2( {x} + 2)^{2} - 5 + 8$

$y = - 2( {x} + 2)^{2} + 3$

The maximum value is 3

The function turns downward, hence the range is:

$y \leqslant 3$

6 0

3 years ago

A container fruits contains 12 3/4 lb of oranges m,8 1/4 lb of bananas and 7 lb of apples what is the total weight in pounds of

notka56 [123]

You just add all of the weight of the contents together. 12 3/4+ 8 1/4 +7, which equals 28 pounds.

3 0

3 years ago

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

6 0

3 years ago