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3 years ago

Find the approximate height of the flagpole.

Mathematics

2 answers:

STatiana [176]3 years ago

8 0

B the answer is 60 feet

DochEvi [55]3 years ago

7 0

Tan(40)=Height÷50

Height
=50(tan(40))
=42 feet (to nearest whole number)

Hence answer is C

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Compute the multifactor productivity measure for each of the weeks shown for production of chocolate bars. What do the productiv

liq [111]

Answer:

Week 1 =3.03

Week 2 =2.99

Week 3=2.9

Week 4=2.83

Step-by-step explanation:

Computation for the multifactor productivity measure for each of the weeks shown for production of chocolate bars.

Using this formula

Week Multifactor productivity measure = (Output / Total Cost)

Where:

Output represent Quantity produced

Total cost represent Labor cost + Material cost + Overhead

Let plug in the formula

Week 1= 30,000/(40*12*6)+[(40*12*6)1.5]+(450*6)

Week 1 =30,000/(2,880)+(2,880*1.5)+2,700

Week 1 =30,000/(2,880+4,320+2,700)

Week 1=30,000/9,900

Week 1 =3.03

Week 2= 33,600/(40*12*7)+[(40*12*7)1.5]+(470*6)

Week 2 =33,600/(3,360)+(3,360*1.5)+2,820

Week 2=33,600/(3,360+5,040+2,820)

Week 2=33,600/11,220

Week 2 =2.99

Week 3= 32,200/(40*12*7)+[(40*12*7)1.5]+(460*6)

Week 3 =32,200/(3,360)+(3,360*1.5)+2,760

Week 3=32,200/(3,360+5,040+2,760)

Week 3=32,200/11,160

Week 3=2.9

Week 4= 35,400/(40*12*8)+[(40*12*8)1.5]+(480*6)

Week 4 =35,400/(3,840)+(3,840*1.5)+2,880

Week 4=35,400/(3,840+5,760+2,880)

Week 4=35,400/12,480

Week 4=2.83

Therefore the multifactor productivity measure for each of the weeks shown for production of chocolate bars is :

Week 1 =3.03

Week 2 =2.99

Week 3=2.9

Week 4=2.83

4 0

2 years ago

A scale drawing of a square has a side length of 6 millimeters. The drawing has a scale of 1 mm : 7 km. Find the actual perimete

zvonat [6]

Answer:

The real perimeter is:

$P= 24 mm *\frac{7 km}{1mm}= 168 km$

And the area would be:

$P = 36mm^2 *\frac{(7km)^2}{(1mm)^2} = 1764 km^2$

Step-by-step explanation:

For this case we know one side of a square given 6mm.

We can begin calculating the perimeter for the scale figure with this formula:

$P = 4x = 4*6 mm =24 mm$

The area for this case is given by:

$A = x^2 = 6mm^2 = 36 mm^2$

And we know that the drawing has a scale of 1 mm : 7 km and if we convert the perimeter to km we got:

$P= 24 mm *\frac{7 km}{1mm}= 168 km$

And the area would be:

$P = 36mm^2 *\frac{(7km)^2}{(1mm)^2} = 1764 km^2$

6 0

3 years ago

What is the value of y in the given triangle?

kondor19780726 [428]

Answer:

Please post a picture of the given triangle.

Step-by-step explanation:

4 0

3 years ago

quadrilateral paid is a rectangle whose diagonals have the endpoints p(-3, -2)i(4, -7) and a(4, -2)d(-3, -7). find the diagonals

julia-pushkina [17]

Answer:

The intersection point of diagonals is (0.5,-4.5).

Step-by-step explanation:

The end points of first diagonal are p(-3, -2) and i(4, -7).

The end points of second diagonal are a(4, -2) and d(-3, -7).

If a line passing through two points, then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The equation of first diagonal is

$y-(-2)=\frac{-7-(-2)}{4-(-3)}(x-(-3))$

$y+2=\frac{-5}{7}(x+3)$

$7y+14=-5(x+3)$

$7y+14=-5x-15$

$5x+7y=-29$ .... (1)

The equation of second diagonal is

$y-(-2)=\frac{-7-(-2)}{-3-4}(x-4)$

$y+2=\frac{-5}{-7}(x-4)$

$7y+14=5(x-4)$

$7y+14=5x-20$

$5x-7y=34$ .... (2)

Add equation (1) and (2),

$10x=5$

$x=0.5$

Put this value in (1).

$5(0.5)+7y=-29$

$y=-4.5$

The intersection point of diagonals is (0.5,-4.5).

3 0

3 years ago

Study this table. x y –3 –2 –2 0 0 4 4 12 Which best describes the function represented by the data in the table? linear with a

ss7ja [257]

we can do this by 2 ways

1- by plotting the points on graph and then tracing the points to get shape,

for linear, we will get straight line

for quadratic, we will get parabola

in this case, it is linear as we get a straight line

2- by solving for values of x and y

consider standard linear equation y = mx +c where m is slope and c is constant

by putting given values of x and y we get

y + 2x = 4(answer)

if we consider standard parabola equation

y^2 = 4ax

this equation is not true for given points

7 0

3 years ago

Read 2 more answers