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3 years ago

12. A photocopier enlarges by a ratio of 7: 4. A picture

Mathematics

1 answer:

Pavlova-9 [17]3 years ago

7 0

Answer: 2 enlargments

Step-by-step explanation:

Use proportion for each enlargment you do

Side a=6cm Side b=4cm

6:4=x:7 4:4=b1:7

7*6=4a1 b1=28/4

a1=42/4 b1=7 after first enlrge

a1=10.5 after first enlarge 7:4=b2:7

10.5:4=a2:7 b2=49/4

a2=73.5/4 xb2=12.5 after 2nd enlarge

a2=18.375 after 2nd enlarge

12.5:4=b3:7

18,375:4=a3:7 b3=21.875 cm after 3rd enlargement

a3=32.15after 3rd enlargement

Values after 3rd enlargement are higher than dimension given for picture wall

30 cm by 20 cm

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4. Here is a linear equation: y=1/4 x + 5/4

4vir4ik [10]

Step-by-step explanation:

"Solutions to the equation" just means that they are points on the line. To find out if these two points land on this line, plug each one in, like this:

1.5 = (1/4)(1) + (5/4)

1.5 = (1/4) + (5/4)

1.5 = (6/4)

1.5 = 1.5

Since the expression is true, this point is on the line.

Do the same process for the second point (remember a point is formatted (x,y)) and see if it is also a point on the line.

To find the x-intercept, simply plug in 0 for y and see what you get. It should look like (x,0).

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2 years ago

Which of all the following inequalities is true for all real values of x?

siniylev [52]

Answer:

C. 1/(x^2 +1) > 0

Step-by-step explanation:

The cube of a negative number is negative, eliminating choices B and D for certain negative values of x.

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The value 1/(x^2+1) is positive everywhere, so that is the expression you're looking for.

1/(x^2 +1) > 0

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3 years ago

Each leg of a 45 45 90 triangle has a length of 6 what is the length of its hypotenuse

Reika [66]

Let x = the length of the triangle's hypotenuse
using proportions, this is what the equation looks like:

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3 years ago

A 1.5-mm layer of paint is applied to one side of the following surface. Find the approximate volume of paint needed. Assume tha

Mariulka [41]

Answer:

V = 63π / 200 m^3

Step-by-step explanation:

Given:

- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:

y = √(42*x - x^2)

- The surface is coated with paint with uniform layer thickness t = 1.5 mm

Find:

The volume of paint needed

Solution:

- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:

$S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx$

- The derivative of the function f'(x) is as follows:

$f'(x) = \frac{21-x}{\sqrt{42x-x^2} }$

- The square of derivative of f(x) is:

$f'(x)^2 = \frac{(21-x)^2}{42x-x^2 }$

- Now use the surface area formula:

$S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi$

- The Volume of the pain coating is:

V = S*t

V = 210*π*3/2000

V = 63π / 200 m^3

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3 years ago

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If the length is 10 ft the width is 4 feet and the height is 8 feet what is the surface area

astraxan [27]

if you're solving for the surface area of a rectangular prism then it's 304 :D

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3 years ago