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2 years ago

What is the scale factor if a 4" by 10" photograph is enlarged to a poster that is 1 ft. by 2.5 ft.?

Mathematics

1 answer:

Lelechka [254]2 years ago

7 0

Answer:

sorry I can't help you with this answer

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A and B are supplementary angles. If A = (5x + 27)° and B = (5x – 7)º, then find the measure of angle A.

vazorg [7]

Answer:

x=16

(5x + 27)° =5(16)+27=80+27=107

(5x – 7)º= 5(16)-7=80-7=73

Step-by-step explanation

supplementary angle=180

(5x + 27)° + (5x – 7)º=180

5x+27+5x-7=180

10x+20=180

10x=180-20

10x=160

x=160/10

x=16

(5x + 27)° =5(16)+27=80+27=107

(5x – 7)º= 5(16)-7=80-7=73

107+73=180

3 0

2 years ago

Find the total surface area of this square based

MrRissso [65]

Answer:

300ft^2

Step-by-step explanation:

Total surface area = Area of base + (1/2) x perimeter of base x slant height

Area of base = length² = 10 x 10 = 100

Perimeter of base = 2 ( length + breadth)

2 x 20 = 40

slant height = 10

100 + (1/2) x 40 x 10 = 300 ft^2

7 0

2 years ago

Determine whether the ratios form a proportion.

-BARSIC- [3]

Answer: Your answer is: A) Yes, Proportional

Step-by-step explanation:

Hope this helped : )

3 0

2 years ago

Read 2 more answers

Gravel is being dumped from a conveyor belt at a rate of 20 ft3 /min and its coarseness is such that it forms a pile in the shap

pantera1 [17]

Answer:

The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

Step-by-step explanation:

Given that :

Gravel is being dumped from a conveyor belt at a rate of 20 ft³ /min

i.e $\dfrac{dV}{dt}= 20 \ ft^3/min$

we know that radius r is always twice the diameter d

i.e d = 2r

Given that :

the shape of a cone whose base diameter and height are always equal.

then d = h = 2r

h = 2r

r = h/2

The volume of a cone can be given by the formula:

$V = \dfrac{\pi r^2 h}{3}$

$V = \dfrac{\pi (h/2)^2 h}{3}$

$V = \dfrac{1}{12} \pi h^3$

$V = \dfrac{ \pi h^3}{12}$

Taking the differentiation of volume V with respect to time t; we have:

$\dfrac{dV}{dt }= (\dfrac{d}{dh}(\dfrac{\pi h^3}{12})) \times \dfrac{dh}{dt}$

$\dfrac{dV}{dt }= (\dfrac{\pi h^2}{4} ) \times \dfrac{dh}{dt}$

we know that:

$\dfrac{dV}{dt}= 20 \ ft^3/min$

So;we have:

$20= (\dfrac{\pi (15)^2}{4} ) \times \dfrac{dh}{dt}$

$20= 56.25 \pi \times \dfrac{dh}{dt}$

$\mathbf{\dfrac{dh}{dt}= \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

8 0

3 years ago

If 1/2 gallon of paint covers 1/12 of a wall then how much paint is needed for the entire wall

netineya [11]

1/2 gallon of paint covers 1/12 of a wall , 12 times as much paint will cover 12 times the area you can solve this by 12*1/2=6.

5 0

3 years ago