Ask question

Ask question

All categories

2 years ago

7

Rotations. Please help me

1 answer:

densk [106]2 years ago

8 0

See the attached picture.

You might be interested in

What types of solutions does 6x^2 - 20x + 1 have?

elena55 [62]

Answer:

2 real solutions

Step-by-step explanation:

We can use the determinant, which says that for a quadratic of the form ax² + bx + c, we can determine what kind of solutions it has by looking at the determinant of the form:

b² - 4ac

If b² - 4ac > 0, then there are 2 real solutions. If b² - 4ac = 0, then there is 1 real solution. If b² - 4ac < 0, then there are 2 imaginary solutions.

Here, a = 6, b = -20, and c = 1. So, plug these into the determinant formula:

b² - 4ac

(-20)² - 4 * 6 * 1 = 400 - 24 = 376

Since 376 is clearly greater than 0, we know this quadratic has 2 real solutions.

<em>~ an aesthetics lover</em>

3 0

2 years ago

If y varies inversely with x<br> and y is 10 when x is 5,<br> then what is y when x is 2?

Jobisdone [24]

9514 1404 393

Answer:

25

Step-by-step explanation:

Inverse variation means ...

y = k/x

For the given values, we can find k to be ...

k = xy = (5)(10) = 50

Then for x=2, we have ...

y = 50/x

y = 50/2

y = 25 . . . when x=2

6 0

3 years ago

Find the product of (√3-5) and (√3+2)

beks73 [17]

Step-by-step explanation:

check it.. i have solved it for you.

3 0

2 years ago

Let f(x) = cx^k be a power function such that f(13) is four times the size of f(1). What is the power k?

Ivenika [448]

Answer:

0.540

Step-by-step explanation:

Hi there,

To get started in order to solve this, please recall the property of logarithms.

This question is a bit tricky, but doable. Best way to solve this is first compare the two functions:

$f(x) = cx^{k} \\ f(13)=4f(1)$

Now, let's see what the function gives at 13 and 1:

$f(13)=c13^{k}\\f(1)=c1^{k}$

Something to recognize is that 1 to the power of <em>anything</em> is just 1, so f(1) is reduced to:

$f(1)=c$ We are making progress!

Next, we can set both f(13) forms equivalent to each other, watch this:

$f(13)=f(13) = 4f(1)$ but we know what f(1) is equal to, and let's substitute our knowns:

$c13^{k}=4c1^{k} = 4c$ the c constant on both left and right side cancel out:

$13^{k}=4$ Now, take the logarithm form of both side of the equation. I used natural log, but you can also use common logs:

$ln(13^{k})=ln(4)$ ⇒ $k*ln(13)=ln(4)$ this was performed using <em>the power log rule. </em>

Isolate k:

$k=\frac{ln(4)}{ln(13)} =0.540$ using a calculator.

If you liked this solution, hit Thanks or give a rating!

thanks,

8 0

2 years ago

The question is "find the lowest common multiple of 4 and 6"<br> with step by step explaination

Tcecarenko [31]

Answer:

12

Step-by-step explanation:

4s multiples:

4,8,12,16,20,24

6s multiples:

6,12,18,24,30,36

lowest number that is a common multiple between both 4 and 6:

12

4 0

2 years ago

Read 2 more answers