All categories

3 years ago

Ann and Pat have 3 cupcakes to share. They cut each cupcake into 2 equal parts.

Mathematics

2 answers:

andrew11 [14]3 years ago

6 0

3/1 divide by 1/2 =6
Remember when you divide fractions you keep the first number the same , change the division sign to multiplication sign and flip the second fraction like this
3/1 x 2/1=6

Rama09 [41]3 years ago

3 0

Answer:

Step-by-step explanation:

3+3 maybe :) .

You might be interested in

Is there a series of rigid transformations that could map A

MaRussiya [10]

Answer:

Yes, A KLP can be reflected across the line containing KP and then translated so that Pis mapped to M.

Step-by-step explanation:

The figure shows two congruent by HA theorem (they have congruent hypotenuses and a pair of congruent angles adjacent to the hypotenuses) triangles KLP and QNM.

A rigid transformation is a transformation which preserves lengths. Reflection, rotation and translation are rigit transformations.

If you reflect triangle KLP across the leg KP and translate it up so that point P coincides with point M , then the image of triangle KLP after these transformations will be triangle QNM.

3 0

3 years ago

Read 2 more answers

Show work please.<br><br> solve system of equations using matrices.

nadya68 [22]

Answer:

(t, t -1, t)

Step-by-step explanation:

You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called. I think it refers to infinite many solutions). Here's how it works:

Set up your matrix:

$\left[\begin{array}{ccc}1&-2&1\\2&-1&-1\\\end{array}\right] \left[\begin{array}{ccc}2\\1\\\end{array}\right]$

You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left. Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0. Multiplying in a -2 to the top row gives you:

$\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right]$

Then add, keeping the first row the same and changing the second to reflect the addition:

$\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right]$

The second equation is this now:

3y - 3z = -3. Solving for y gives you y = z - 1. Let's let z = t (some random real number that will make the system true. Any number will work. I'll show you at the end. Just bear with me...)

lf z = t, and if y = z - 1, then y = t - 1. So far we have that y = t - 1 and z = t. Now we solve for x:

From the first equation in the original system,

x - 2y + z = 2. Subbing in t - 1 for y and t for z:

x - 2(t - 1) + t = 2. Simplify to get

x - 2t + 2 + t = 2 and x - t = 0, and x = t. So the solution set is (t, t - 1, t). Picking a random value for t of, let's say 2, sub that in and make sure it works. If:

x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2. Check it: 2 - 4 + 4 = 2 and 2 = 2. You could pick any value for t and it will work.

6 0

3 years ago

Assume that A varies directly with z. If A = 30 when z = 5, find A when z = 9.

pishuonlain [190]

If A varies directly with z, and A is 30 and z is 5, then when z is 1, A is 6.
z: 5 ÷ 5 = 1
A: 30 ÷ 5 = 6
Then, when z is 1, A is 6
So, when z is 9, A is...
z: 1 × 9 = 9
A: 6 × 9 = 54

A is 54

6 0

3 years ago

Read 2 more answers

Is my answer correct?

GalinKa [24]

Your answer is correct :)

3 0

3 years ago

Em 1 Graph the system of equations. { x−y=6 ,4x+y=4

muminat

Graphing the system of equations is shown in figure attached.

Solution set is (2,-4). The lines will intersect at (2,-4)