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4 years ago

How can you use scale models to solve problems

Mathematics

1 answer:

tino4ka555 [31]4 years ago

6 0

<span>Scale drawings make it easy to see large things on paper, like a house or road</span>

Read more about arc and chords at:

brainly.com/question/25871159

3 0

2 years ago

In 2005 there were 1352 students in 2006 there were 1014 students work out the decrease in the number of students

Eduardwww [97]

I’m. It sure but I need points so sorry

6 0

3 years ago

Statistics:

Dima020 [189]

Answer:

Step-by-step explanation:

a) sub in 600 for 6

600-25/40 =14.375

b) switch x and y

x=y-25/40

Solve for y

40x=y-25

y=40x+25

c)Plug in 2.5 for z(x)

2.5= x-25/40

100=x-25

x=125

8 0

3 years ago

The date of death for a widow was 2019. If the estate was valued at $2,186,000 and the estate was taxed at 40 percent, what was

Airida [17]

Answer:

Value of tax liability will be zero

Step-by-step explanation:

Given:

Total estate value = $2,186,000

Estate tax = 40 %

Find:

Heir's tax liability = ?

Computation:

Total taxable estate = Total estate value - Estate Exemption

Note: According to Exhibit 19-6 for Estate Exemption, Estate Exemption is 11.4 million.

We know that Estate Exemption is higher than total estate value, therefore will be no tax liability on the successor, so the value of tax liability will be zero.

7 0

3 years ago

A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a

Rashid [163]

Answer:

Length of original rectangle: 11 units.

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

$\text{Perimeter of new rectangle}=20$

Step-by-step explanation:

Let x represent width of the original rectangle.

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be $x+6$ .

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be $x-2$ .

The length of new rectangle would be $x+6-4=x+2$ .

The area of new rectangle would be $(x+2)(x-2)$ .

Now we will equate area of new rectangle with 21 and solve for x as:

$(x+2)(x-2)=21$

Applying difference of squares, we will get:

$x^2-2^2=21$

$x^2-4=21$

$x^2-4+4=21+4$

$x^2=25$

Since width cannot be negative, so we will take positive square root of both sides.

$\sqrt{x^2}=\sqrt{25}$

$x=5$

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be $x+6\Rightarrow x+5=11$ .

Therefore, the length of original rectangle is 11 units.

$\text{Area of original rectangle}=5\times 11$

$\text{Area of original rectangle}=55$

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

We know that perimeter of rectangle is two times the sum of length and width.

$\text{Perimeter of new rectangle}=2((x+2)+(x-2))$

$\text{Perimeter of new rectangle}=2((5+2)+(5-2))$

$\text{Perimeter of new rectangle}=2(7+3)$

$\text{Perimeter of new rectangle}=2(10)$

$\text{Perimeter of new rectangle}=20$

Therefore, the perimeter of the new rectangle is 20 units.

7 0

4 years ago