3. Kyle starts with $15.00 and saves $3.50

If you want to prove that the diagonals of a parallelogram bisect each other using coordinate geometry, how would you place the

katen-ka-za [31]

Answer:

In general you can choose the vertices at any arbitrary points but for easier computations and calculations we can choose 1 vertex at origin with co-ordinates $(0,0)$ and it's adjacent vertex either on x-axis with co-ordinates $(x,0)$ or on y-axis with ordinates $(0,y)$

Thus the coordinates of vertices become

5 0

3 years ago

Steve made $50 raking leaves and $75 mowing lawns. He gave 60% of his earnings to his mother. How much did Steve give to his mot

Lena [83]

Answer:

$75

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Please help, I dont know how to do this

Mrrafil [7]

Answer:

$\frac{y-2}{4}$

Step-by-step explanation:

Let's start by factoring everything:

$\dfrac{\frac{y^2+y}{y^2-2y}}{\frac{4y+4}{y^2-4y+4}}=$

$\dfrac{\frac{y(y+1)}{y(y-2)}}{\frac{4(y+1)}{(y-2)^2}}=$

Now, you can cancel out some terms:

$\dfrac{\frac{y+1}{y-2}}{\frac{4(y+1)}{(y-2)^2}}=$

Now, when you divide a fraction by another fraction, that is the same as multiplying the first fraction by the reciprocal of the second one:

$\frac{y+1}{y-2}\cdot \frac{(y-2)^2)}{4(y+1)}=$

$\frac{y-2}{4}$

Hope this helps!

4 0

4 years ago

In the diagram below of triangle PQR, S is the midpoint of PR and T is the

Nina [5.8K]

Answer:

ST = 4

Step-by-step explanation:

A segment joining the midpoints of 2 sides of a triangle is half the length of the third side.

ST = $\frac{1}{2}$ PQ substitute values

- 32 + 9x = $\frac{1}{2}$ (- 5x + 28) ← multiply both sides by 2 to clear the fraction

- 64 + 18x = - 5x + 28 (add 5x to both sides )

- 64 + 23x = 28 ( add 64 to both sides )

23x = 92 ( divide both sides by 23 )

x = 4

Then

ST = - 32 + 9x = - 32 + 9(4) = - 32 + 36 = 4

5 0

3 years ago

Can you help me with number 6? <br> Confused abit <br> Please

Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.

To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.

First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is

$\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}$

Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is

$\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}$

Third question:

To get exactly one six, it can either be the first, second or third roll.

In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.

In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus

$\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}$

And since there are three of these combinations, The answer is

$3\frac{5^2}{6^3}$

Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is

$1 - \frac{5^3}{6^3}$

4 0

3 years ago