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3 years ago

A square with a side length of 3 units is dilated with a scale factor of 2.

Mathematics

1 answer:

Anna [14]3 years ago

3 0

<h3> - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - </h3>

➷Simply multiply by 2:

3 x 2 = 6

It would be 6 units

➶Hope This Helps You!

➶Good Luck :)

➶Have A Great Day ^-^

↬ Hannah ♡

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Prove<br>that:<br>cos^3A Cos3A + sin^3A.sin 3A=cos^3 2A

aalyn [17]

Answer:

3 0

3 years ago

Which are steps that could be used to solve 0 = 9(x2 + 6x) – 18 by completing the square? Check all that apply.

jeyben [28]

18 + 81 = 9(x²<span> + 6x + 9)
</span><span>11 = (x + 3)</span>²

When we are completing the square, we are going to move the value of c across the equals. We will do that by adding, and end up with

18=9(x²+6x)

We take the value of b (the coefficient of x), divide it by 2 and square it:
(6/2)²=3²=9

This is the value that completes the square. However, since the entire square is multiplied by 9, this value must be multiplied by 9 before we can add it across the equals:
18+9(9) = 9(x²+6x+9)
18+81=9(x²+6x+9)
99=9(x²+6x+9)

Dividing both sides by 9, we have:
11=x²+6x+9

11=(x+3)²

8 0

3 years ago

Read 2 more answers

Math:

Dahasolnce [82]

Answer:

a) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$ , b) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

Step-by-step explanation:

a) The equation must be rearranged into a form with one fundamental trigonometric function first:

$\sqrt{3}\cdot \csc x - 2 = 0$

$\sqrt{3} \cdot \left(\frac{1}{\sin x} \right) - 2 = 0$

$\sqrt{3} - 2\cdot \sin x = 0$

$\sin x = \frac{\sqrt{3}}{2}$

$x = \sin^{-1} \frac{\sqrt{3}}{2}$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

b) The equation must be simplified first:

$\cos x + 1 = - \cos x$

$2\cdot \cos x = -1$

$\cos x = -\frac{1}{2}$

$x = \cos^{-1} \left(-\frac{1}{2} \right)$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

7 0

3 years ago

I need help with this question asap!!I will give brainliest!!!please and thank you

pogonyaev

Answer:

Step-by-step explanation:

(4,6) and (20,14) are points on the line.

slope of line = (14-6)/(20-4) = 8/16= 1/2

point-slope equation for line of slope 1/2 that passes through (6,4):

y-4 = (½)(x-6)

in slope-intercept form:

y = ½x + 1

y-intercept = 1

6 0

2 years ago

A boat sails 285 miles south and then 132 miles west. What is the magnitude of the boats resultant vector?

Delicious77 [7]

The magnitude of the boats resultant vector is 314.1 mi

<h3>What is a vector?</h3>

A vector is a physical quantity that has both magnitude and direction.

<h3>What is a resultant vector?</h3>

A resultant vector is the sum of two or more vectors.

<h3>How to find the boats resultant vector?</h3>

Since the boat sails 285 miles south and then 132 miles west, we have that its first direction vector is r = (285 mi)j. Also, its direction vector west is r' = -(132 mi)i

So, the resultant vector R = r + r'

= (285 mi)j + (132 mi)i

= (132 mi)i + (285 mi)j

So, the magnitude of the resultant vector is R = √(r² + r'²)

So, substituting thevalues of the variables into the equation, we have

R = √(r² + r'²)

R = √((285 mi)² + (132 mi)²)

R = √(81225 mi² + 17424 mi²)

R = √(98649 mi²)

R = 314.08 mi

R ≅ 314.1 mi

So, the resultant vector is 314.1 mi

Learn more about magnitude of resultant vector here:

brainly.com/question/28047791

#SPJ1

8 0

1 year ago