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Maslowich
3 years ago
5

Explain how an enlargement or a reduction in the dimensions of a building would cause a change in the scale factor.

Mathematics
2 answers:
sleet_krkn [62]3 years ago
8 0

The scale factor represents a proportional relationship between the scale drawing and the building’s dimensions. If the dimensions change, then the scale factor must also change to preserve the relationship.

swat323 years ago
5 0

Answer:

It would create an enlarged figure.

Explanation:

A scale factor of  

1

results in a figure of equal dimensions or an isometric figure.

A scale factor of less than  

1

results in a figure of reduced dimensions .

A scale factor of greater than  

1

would result in a figure of enlarged dimensions.

As  

4

3

>

1

, it would create an enlarged figure.

Step-by-step explanation:


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I need help with this equation please!<br><br> 15- w/4 = 28
Luda [366]

Answer:

w= -52

Step-by-step explanation:

We simplify the equation to the form, which is simple to understand  

15-w/4=28

Simplifying:

15-0.25w=28

We move all terms containing w to the left and all other terms to the right.  

-0.25w=+28-15

We simplify left and right side of the equation.  

-0.25w=+13

We divide both sides of the equation by -0.25 to get w.  

w= -52

Hope it Helps :) !!

8 0
3 years ago
What is the place value of 6 in the number 6,029?​
Feliz [49]

Answer: 6 on the extreme left has the place value 6000; the next 6 has the value 600; the next, 60; and the last, 6. The numeral for every whole number stands for a sum. 364 = 3 Hundreds + 6 Tens + 4 Ones. Hope this helps!

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Hello again! This is another Calculus question to be explained.
podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

Functions

  • Function Notation
  • Exponential Property [Rewrite]:                                                                   \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Property [Root Rewrite]:                                                           \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                 \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

We are given the following and are trying to find the second derivative at <em>x</em> = 2:

\displaystyle f(2) = 2

\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}

When we differentiate this, we must follow the Chain Rule:                             \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]

Use the Basic Power Rule:

\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]

Simplifying it, we have:

\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]

We can rewrite the 2nd derivative using exponential rules:

\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}

To evaluate the 2nd derivative at <em>x</em> = 2, simply substitute in <em>x</em> = 2 and the value f(2) = 2 into it:

\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}

When we evaluate this using order of operations, we should obtain our answer:

\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

5 0
2 years ago
Please I need help of this one :)
Harlamova29_29 [7]

Answer:

128.605 m²

Step-by-step explanation:

Find the areas of the top triangle, and the bottom triangle and add them together.

The top triangle consists of 2 smaller triangles.  Look at the larger one, the height is 4.93 and the base is 17.  The area is...

A = (1/2)(17)(4.93)

   A = 41.905 m²

The bottom triangle consists of 2 smaller triangles.  Look at the larger one, the height is 10.2 and the base is 17.   The area is...

A = (1/2)(17)(10.2)

  A = 86.7 m²

The total area is 41.905 m² + 86.7 m² = 128.605 m²

5 0
3 years ago
Please help fast!!!!!!!!!!!!!!
denis-greek [22]

Answer:

The third option, -1/2(30) = -15

Step-by-step explanation:

If it decreases by 1/2, we know it has to be -1/2, and the days we are given are 30. We need to multiply -1/2 by 30 in order to find the water level after 30 days, which would be -15 (inches?) I hope this helped.

5 0
2 years ago
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