How do you solve for mass if KE=0.5mv2

What is the difference of the two polynomials?<br><br> (92+8)–(22+3)

stich3 [128]

Answer:

75

Step-by-step explanation:

(92+8)-(22+3)

Add the terms inside of the parentheses because of PEMDAS (Parenthesis, Exponent, Multiplication, Division, Addition, Subtraction)

(100)-(25)

Then subtract 25 from 100

75

8 0

3 years ago

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How is it possible to simplify 2 divided by (the square root of 6)? My calculator says that it is equal to (the square root of 6

Sauron [17]

Answer:

$\boxed{\frac{\sqrt{6}}{3}}$

Step-by-step explanation:

Multiply the fraction by $\frac{\sqrt{6}}{\sqrt{6}}$

$\frac{2}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{2\sqrt{6}}{6}\\$

Cancel the 2s

$\frac{2\sqrt{6}}{6}= \frac{\sqrt{6}}{3}$

∴ $\boxed{\frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}}$

6 0

3 years ago

Fiona, an avid food blogger posted a total 216 recipes in 36 weeks. If she posted them at a constant rate, how many recipes did

Cloud [144]

Answer:

90

Step-by-step explanation:

216/36= 6

6 times 15=90

Hope this helped

:)

5 0

3 years ago

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Quadrilateral ABCD has vertices A(-2,4), B(-3.-3), C(-5,-2), D(-5,2). Which of the following are the coordinates of A’ after a d

Gala2k [10]

<h2>Answer:</h2>

$\boxed{A'(-5,10)}$

<h2>Step-by-step explanation:</h2>

Dilation is when you stretch something by the same amount in two perpendicular directions. In other words, dilation changes size, not overall shape. Dilation factors greater than one makes the shape greater while dilation factors less than one makes the shape smaller. The scale factor in this problem is:

$2\frac{1}{2}$

Which is a mixed fraction. This can be transformed into an improper fraction as follows:

$2\frac{1}{2}=2+\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2}$

So the scale factor is $\frac{5}{2}$ . Since the center of dilation is the origin, we just need to multiply each vertex of the quadrilateral by the scale factor. Thus, for A we have:

$A'(-2\times \frac{5}{2},4\times \frac{5}{2})=\boxed{A'(-5,10)}$

3 0

4 years ago

An experiment consists of tossing a die and then flipping a coin once if the number on the die is even. If the number on the die

Mice21 [21]

Answer:

$\Omega = \{ 1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T,\\5HH , 5HT, 5TH, 5TT, 6H, 6T \}$

The probability is 3/12. The third option is correct.

Step-by-step explanation:

The sample space is

$\Omega = \{ 1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T,\\5HH , 5HT, 5TH, 5TT, 6H, 6T \}$

Note that this sample space is not equally probable.

The probability of getting a given number followed is the probability of getting an even number from the 6 numbers (3/6) multiplied by the probability of getting a head after getting that even number, that is 1/2, because is equally probable to get heads or tails from one single coin toss (note that we are assuming that the dice was even, thats why there is a single coin toss).

Therefore, the probability of getting an even number and a head is

P( D in {2,4,6} , H = 1) = P(D in {2,4,6}) * P(H=1 | D in {2,4,6}) = 3/6 * 1/2 = 3/12.

5 0

3 years ago