Ask question

Ask question

All categories

3 years ago

11

Point N ( , 2 9) is on RS. A translation moves point N to its image, N'(6 3 , ). What is the distance, in units, between any poi

nt on RS and its image? A 10 B 40 C 52 D 130

1 answer:

KatRina [158]3 years ago

7 0

Hello!

Distance between two points (2,9) and (6,3) -

<span>1. Take the Absolute Difference between the X points 2 and 6. The Absolute Difference is 4. </span>
<span>2. Square 4 = 16 </span>
<span>3. Take the Absolute Difference between the Y points 9 and 3. The absolute Difference is 12. </span>
<span>4. Square 12 = 144 </span>
<span>5 Add 144 + 16 = 160 </span>
<span>6. Take the Square root of 160 </span>

<span>12.6491106407

Hope this Helps at least! :)</span>

You might be interested in

A³b² a²b simplify the following expression

liberstina [14]

Answer:

$a^{5}$ $b^{3}$

Step-by-step explanation:

The law of indices can be used to simplify mathematical expressions involving arithmetical operation on variables with powers.

$a^{m}$ x $a^{n}$ = $a^{(m+n)}$

Thus, the given expression can be simplified as follows:

a³b² a²b = a³ x a² x b² x $b^{1}$

= $a^{3+2}$ x $b^{2+1}$

= $a^{5}$ $b^{3}$

Thus,

a³b² a²b = $a^{5}$ $b^{3}$

7 0

2 years ago

20% of 2kg ( in g). <br> Please help

Svetlanka [38]

Answer:

${ \tt{ = 20\% \times 2}} \\ = 0.4 \: kg \\ { \tt{ = 400 \: grams}}$

3 0

3 years ago

Read 2 more answers

There are 2 each shape is 1 whole. what fraction greater than 1 names the parts that are shaded?

lilavasa [31]

It's three because of you have two anything bigger than two is three

6 0

3 years ago

Help me find the error <br> plz i beg

Annette [7]

Answer:

you have to divide by 4 not subtracted.

Step-by-step explanation:

5 0

3 years ago

For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va

Black_prince [1.1K]

Answer:

3 solutions:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is $tan(\frac{\theta}{2})$ so let's focus on that part of the equation first.

We know that:

$tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

therefore:

$cos(\frac{\theta}{2})\neq0$

so we need to find the angles that will make the cos function equal to zero. So we get:

$cos(\frac{\theta}{2})=0$

$\frac{\theta}{2}=cos^{-1}(0)$

$\frac{\theta}{2}=\frac{\pi}{2}+\pi n$

or

$\theta=\pi+2\pi n$

we can now start plugging values in for n:

$\theta=\pi+2\pi (0)=\pi$

if we plugged any value greater than 0, we would end up with an angle that is greater than $2\pi$ so, that's the only angle we cannot include in our answer set, so:

$\theta\neq \pi$

having said this, we can now start solving the equation:

$tan(\frac{\theta}{2})=sin(\theta)$

we can start solving this equation by using the half angle formula, such a formula tells us the following:

$tan(\frac{\theta}{2})=\frac{1-cos(\theta)}{sin(\theta)}$

so we can substitute it into our equation:

$\frac{1-cos(\theta)}{sin(\theta)}=sin(\theta)$

we can now multiply both sides of the equation by $sin(\theta)$

so we get:

$1-cos(\theta)=sin^{2}(\theta)$

we can use the pythagorean identity to rewrite $sin^{2}(\theta)$ in terms of cos:

$sin^{2}(\theta)=1-cos^{2}(\theta)$

so we get:

$1-cos(\theta)=1-cos^{2}(\theta)$

we can subtract a 1 from both sides of the equation so we end up with:

$-cos(\theta)=-cos^{2}(\theta)$

and we can now add $cos^{2}(\theta)$

to both sides of the equation so we get:

$cos^{2}(\theta)-cos(\theta)=0$

and we can solve this equation by factoring. We can factor $cos(\theta)$ to get:

$cos(\theta)(cos(\theta)-1)=0$

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

$cos(\theta)=0$

$\theta=cos^{-1}(0)$

$\theta={\frac{\pi}{2}, \frac{3\pi}{2}}$

Equation 2:

$cos(\theta)-1=0$

we add a 1 to both sides of the equation so we get:

$cos(\theta)=1$

$\theta=cos^{-1}(1)$

$\theta=0$

so we end up with three answers to this equation:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

7 0

2 years ago