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olasank [31]
3 years ago
10

A contractors crew can frame 3 houses in a week. How long will it take them to frame 54 houses if they frame the same number eac

h week.
Mathematics
1 answer:
BartSMP [9]3 years ago
5 0
I dun no good luck though
You might be interested in
A roof is shaped like a cone with a diameter of 12 feet. One bundle of shingles covers 32 square feet. How many bundles should y
max2010maxim [7]

Answer:

113.04 (6 + g) / 32

Step-by-step explanation:

We know that the total area of a cone is:

A = Pi * r * (r + g), let g be the generatrix.

The radius is half the diameter, therefore, 12/2 = 6 feet.

Replacing:

A = 3.14 * (6) * (6 + g) = 113.04 (6 + g) square feet.

To know the total number of tiles package, we must divide the area of the cone by the area occupied by a package. So:

113.04 (6 + g) / 32

The problem statement does not have a height or the value of the generatrix therefore it is not possible to calculate the number of packages.

3 0
3 years ago
For a triangle $XYZ$, we use $[XYZ]$ to denote its area. Let $ABCD$ be a square with side length $1$. Points $E$ and $F$ lie on
nata0808 [166]

An algebraic equation enables the expression of equality between variable expressions

\underline{The \ value \ of \ [AEF] \ is \ \dfrac{4}{9}}

The reason the above value is correct is given as follows:

The given parameters are;

The symbol for the area of a triangle ΔXYZ = [XYZ]

The side length of the given square ABCD = 1

The location of point <em>E</em> = Side \overline{BC} on square ABCD

The location of point <em>F</em> = Side \overline{CD} on square ABCD

∠EAF = 45°

The area of ΔCEF, [CEF] = 1/9 (corrected by using a similar online question)

Required:

To find the value of [AEF]

Solution:

The area of a triangle = (1/2) × Base length × Height

Let <em>x</em> = EC, represent the base length of ΔCEF, and let <em>y</em> = CF represent the height of triangle ΔCEF

We get;

The area of a triangle ΔCEF, [CEF] = (1/2)·x·y = x·y/2

The area of ΔCEF, [CEF] = 1/9 (given)

∴ x·y/2 = 1/9

ΔABE:

\overline{BE} = BC - EC = 1 - x

The area of ΔABE, [ABE] = (1/2)×AB ×BE

AB = 1 = The length of the side of the square

The area of ΔABE, [ABE] = (1/2)× 1 × (1 - x) = (1 - x)/2

ΔADF:

\overline{DF} = CD - CF = 1 - y

The area of ΔADF, [ADF] = (1/2)×AD ×DF

AD = 1 = The length of the side of the square

The area of ΔADF, [ADF] = (1/2)× 1 × (1 - y) = (1 - y)/2

The area of ΔAEF, [AEF] = [ABCD] - [ADF] - [ABE] - [CEF]

[ABCD] = Area of the square = 1 × 1

[AEF] = 1 - \dfrac{1 - x}{2} - \dfrac{1 - y}{2} - \dfrac{1}{19}= \dfrac{19 \cdot x + 19 \cdot y - 2}{38}

From \dfrac{x \cdot y}{2} = \dfrac{1}{9}, we have;

x = \dfrac{2}{9 \cdot y}, which gives;

[AEF] =  \dfrac{9 \cdot x + 9 \cdot y - 2}{18}

Area of a triangle = (1/2) × The product of the length of two sides × sin(included angle between the sides)

∴ [AEF] =  (1/2) × \overline{AE} × \overline{FA} × sin(∠EAF)

\overline{AE} = √((1 - x)² + 1), \overline{FA}  = √((1 - y)² + 1)

[AEF] =  (1/2) × √((1 - x)² + 1) × √((1 - y)² + 1) × sin(45°)

Which by using a graphing calculator, gives;

\dfrac{1}{2} \times \sqrt{(1 - x)^2 + 1} \times \sqrt{(1 - y)^2 + 1} \times \dfrac{\sqrt{2} }{2} =  \dfrac{9 \cdot x + 9 \cdot y - 2}{18}

Squaring both sides and plugging in x = \dfrac{2}{9 \cdot y}, gives;

\dfrac{(81 \cdot y^4-180 \cdot y^3 + 200 \cdot y^2 - 40\cdot y +4)\cdot y^2}{324\cdot y^4}  = \dfrac{(81\cdot y^4-36\cdot y^3 + 40\cdot y^2 - 8\cdot y +4)\cdot y^2}{324\cdot y^2}

Subtracting the right hand side from the equation from the left hand side gives;

\dfrac{40\cdot y- 36\cdot y^2 + 8}{81\cdot y} = 0

36·y² - 40·y + 8 = 0

y = \dfrac{40 \pm \sqrt{(-40)^2-4 \times 36\times 8} }{2 \times 36} = \dfrac{5 \pm \sqrt{7} }{9}

[AEF] =  \dfrac{9 \cdot x + 9 \cdot y - 2}{18} = \dfrac{9 \cdot y^2-2 \cdot y + 2}{18 \cdot y}

Plugging in y =  \dfrac{5 + \sqrt{7} }{9} and rationalizing surds gives;

[AEF] =  \dfrac{9 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) ^2-2 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right)  + 2}{18 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) } = \dfrac{\dfrac{40+8\cdot \sqrt{7} }{9} }{10+2\cdot \sqrt{7} } = \dfrac{32}{72} = \dfrac{4}{9}

Therefore;

\underline{[AEF]= \dfrac{4}{9}}

Learn more about the use of algebraic equations here:

brainly.com/question/13345893

6 0
3 years ago
What’s the answer to rounding 5,374 to the nearest hundred
Radda [10]
5,400 is the answer to the question that you have asked
8 0
3 years ago
Read 2 more answers
Carol's Phone Service charges $17.62 for an activation fee plus $3 for every gigabyte (g) used. Which of the following graphs mo
Aleks [24]

Answer:

the one that starts at 0,17.62 and has a slope of 3

Step-by-step explanation:

3 0
4 years ago
) Write the slope-intercept form of the equation of the line described by (−4, −5), parallel to y = −2x − 5.
Helga [31]

Answer:

y = - 2x - 13

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - 2x - 5 ← is in slope- intercept form

with slope m = - 2

Parallel lines have equal slopes, thus

y = - 2x + c ← is the partial equation

To find c substitute (- 4, - 5) into the partial equation

- 5 = 8 + c ⇒ c = - 5 - 8 = - 13

y = - 2x - 13 ← equation of parallel line

7 0
3 years ago
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