All categories

3 years ago

For the dilation, DO, K = (10, 0) → (5, 0), the scale factor is equal to...?

Mathematics

2 answers:

romanna [79]3 years ago

4 0

The scale factor would be 1/2

Katyanochek1 [597]3 years ago

4 0

Answer:

The scale factor is $\frac{1}{2}$ .

Step-by-step explanation:

The given rule of dilation is

$D_{O,K}=(10,0)\rightarrow (5, 0)$

It means the figure is dilated by scale factor K with center at origin.

If a figure dilated by scale factor K center at origin, then the scale factor is

$K=\frac{\text{x-coordinate of image}}{\text{x-coordinate of preimage}}$

The x-coordinate of image is 5 and the x-coordinate of preimage is 10. So, the scale factor is

$K=\frac{5}{10}$

$K=\frac{1}{2}$

Therefore the scale factor is $\frac{1}{2}$ .

You might be interested in

What is the value of X

Sindrei [870]

The answer is: [C]: " 3 " .
____________________________________
Explanation:
__________________________________
4x - 1 = 2x + 5 ; Solve for "x" ;
__________________________________
Subtract "2x" from each side of the equation; and add "1" to each side of the equation:
___________________________________________________
4x - 1 - 2x + 1 = 2x + 5 - 2x + 1 ;
___________________________________________________
2x = 6 ;
_________________________________
Divide EACH side of the equation by "2" ; to isolate "x" on ONE side of the equation; and to solve for "x" ;
_________________________________
2x / 2 = 6 / 2 ;
_________________________________
x = 3 .
_________________________________
The answer is: [C]: " 3 " .
_________________________________

4 0

3 years ago

Match each vector operation with its resultant vector expressed as a linear combination of the unit vectors i and j.

Cloud [144]

Answer:

3u - 2v + w = 69i + 19j.

8u - 6v = 184i + 60j.

7v - 4w = -128i + 62j.

u - 5w = -9i + 37j.

Step-by-step explanation:

Note that there are multiple ways to denote a vector. For example, vector u can be written either in bold typeface "u" or with an arrow above it $\vec{u}$ . This explanation uses both representations.

$\displaystyle \vec{u} = \langle 11, 12\rangle =\left(\begin{array}{c}11 \\12\end{array}\right)$ .

$\displaystyle \vec{v} = \langle -16, 6\rangle= \left(\begin{array}{c}-16 \\6\end{array}\right)$ .

$\displaystyle \vec{w} = \langle 4, -5\rangle=\left(\begin{array}{c}4 \\-5\end{array}\right)$ .

There are two components in each of the three vectors. For example, in vector u, the first component is 11 and the second is 12. When multiplying a vector with a constant, multiply each component by the constant. For example,

$3\;\vec{v} = 3\;\left(\begin{array}{c}11 \\12\end{array}\right) = \left(\begin{array}{c}3\times 11 \\3 \times 12\end{array}\right) = \left(\begin{array}{c}33 \\36\end{array}\right)$ .

So is the case when the constant is negative:

$-2\;\vec{v} = (-2)\; \left(\begin{array}{c}-16 \\6\end{array}\right) =\left(\begin{array}{c}(-2) \times (-16) \\(-2)\times(-6)\end{array}\right) = \left(\begin{array}{c}32 \\12\end{array}\right)$ .

When adding two vectors, add the corresponding components (this phrase comes from Wolfram Mathworld) of each vector. In other words, add the number on the same row to each other. For example, when adding 3u to (-2)v,

$3\;\vec{u} + (-2)\;\vec{v} = \left(\begin{array}{c}33 \\36\end{array}\right) + \left(\begin{array}{c}32 \\12\end{array}\right) = \left(\begin{array}{c}33 + 32 \\36+12\end{array}\right) = \left(\begin{array}{c}65\\48\end{array}\right)$ .

Apply the two rules for the four vector operations.

$\displaystyle \begin{aligned}3\;\vec{u} - 2\;\vec{v} + \vec{w} &= 3\;\left(\begin{array}{c}11 \\12\end{array}\right) + (-2)\;\left(\begin{array}{c}-16 \\6\end{array}\right) + \left(\begin{array}{c}4 \\-5\end{array}\right)\\&= \left(\begin{array}{c}3\times 11 + (-2)\times (-16) + 4\\ 3\times 12 + (-2)\times 6 + (-5) \end{array}\right)\\&=\left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle\end{aligned}$

Rewrite this vector as a linear combination of two unit vectors. The first component 69 will be the coefficient in front of the first unit vector, i. The second component 19 will be the coefficient in front of the second unit vector, j.

$\displaystyle \left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle = 69\;\vec{i} + 19\;\vec{j}$ .

$\displaystyle \begin{aligned}8\;\vec{u} - 6\;\vec{v} &= 8\;\left(\begin{array}{c}11\\12\end{array}\right) + (-6) \;\left(\begin{array}{c}-16\\6\end{array}\right)\\&=\left(\begin{array}{c}88+96\\96 - 36\end{array}\right)\\&= \left(\begin{array}{c}184\\60\end{array}\right)= \langle 184, 60\rangle\\&=184\;\vec{i} + 60\;\vec{j} \end{aligned}$ .

$\displaystyle \begin{aligned}7\;\vec{v} - 4\;\vec{w} &= 7\;\left(\begin{array}{c}-16\\6\end{array}\right) + (-4) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}-112 - 16\\42+20\end{array}\right)\\&= \left(\begin{array}{c}-128\\62\end{array}\right)= \langle -128, 62\rangle\\&=-128\;\vec{i} + 62\;\vec{j} \end{aligned}$ .

$\displaystyle \begin{aligned}\;\vec{u} - 5\;\vec{w} &= \left(\begin{array}{c}11\\12\end{array}\right) + (-5) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}11-20\\12+25\end{array}\right)\\&= \left(\begin{array}{c}-9\\37\end{array}\right)= \langle -9, 37\rangle\\&=-9\;\vec{i} + 37\;\vec{j} \end{aligned}$ .

7 0

3 years ago

What is the output value of (4x+2)(x-5)/2 using input value of x=2

Firdavs [7]

Answer:

-15

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Step-by-step explanation:

<u>Step 1: Define</u>

(4x + 2)(x - 5) / 2

x = 2

<u>Step 2: Evaluate</u>

Substitute: (4 · 2 + 2)(2 - 5) / 2
Parenthesis - Multiply: (8 + 2)(2 - 5) / 2
Parenthesis - Add: 10(2 - 5) / 2
Parenthesis - Subtract: 10(-3) / 2
Multiply: -30 / 2
Divide: -15

6 0

3 years ago

Read 2 more answers

if a sequence is defined recursively by f(0)=4 and f(n+1)= -3f(n+1)=-3f(n)+1 for n≥0, then f(3) is equal to? DESPERATELY NEED HE

muminat

f(3) is equal to -101.

<u>Step-by-step explanation</u>:

The sequence is followed by f(0)=4
The given expression is f(n+1)= -3f(n)+1

Put n=0,

f(0+1)= -3f(0)+1

f(1)= -3(4)+1

f(1)= -12+1 = -11

Put n=1,

f(1+1)= -3f(1)+1

f(2)= -3(-11)+1

f(2)= 33+1 = 34

Put n=2,

f(2+1)= -3f(2)+1

f(3)= -3(34)+1

f(3)= -102+1

f(3)= -101

5 0

3 years ago

1. Evaluate: 5 ÷1 + 3 + 7

Elena-2011 [213]

Answer:

Step-by-step explanation:

5÷1+3+7

PEMDAS

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

5÷1= 5, then 5+3=8, then 8+7= 15

3 0

3 years ago