All categories

3 years ago

OMG OMG OMG PLEASE HELP!!!!!!!!!!!

Mathematics

2 answers:

Leya [2.2K]3 years ago

6 0

Answer:

Step-by-step explanation:

$\frac{3-\sqrt{3} }{3+\sqrt{3} } *\frac{3-\sqrt{3} }{3-\sqrt{3} } \\\\=\frac{(3-\sqrt{3})^{2} }{(3+\sqrt{3} )(3-\sqrt{3} )} \\\\=\frac{9-2*3\sqrt{3}+3 }{9-3}\\\\=\frac{12-6\sqrt{3} }{6} \\\\=2-\sqrt{3}$

krok68 [10]3 years ago

4 0

The answer for that is c

You might be interested in

How to solve 620 divided by 10 to the power of 2

uranmaximum [27]

Answer:

6 1/5

hope this helps!

5 0

3 years ago

Parker has quarters and dimes in his piggy bank he has 4 more dimes than quarters and he has a total of $7.05 in his bank how ma

den301095 [7]

Parker has 23 dimes and 19 quarters.

3 0

3 years ago

Read 2 more answers

Apply each of the four counting/sampling methods (with replacement and with ordering, without replacement and with ordering, wit

dusya [7]

Answer:

Step-by-step explanation:

I will illustrate this solution with a unique birthday party situation between Jane (the celebrant) and her 10 friends.

In the said party , we will assume that Jane only has 5 chocolates to share among her friends

i. Assuming that the 5 chocolates are of the same type, if she doesn't want to give any friend more than one piece of chocolate, the situation here is said to be WITHOUT REPLACEMENT & UN-ORDERED

n = 10 and k = 5

Total number of possible combinations becomes,

$(\left {n} \atop {k}} \right. )=(\left {10} \atop {5}} \right. )=252$

ii. Assuming that the 5 pieces of chocolate are of the same type and she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & UN-ORDERED

n = 10 and k = 5

Total number of possible combinations becomes,

$(\left {n+k-1} \atop {k}} \right. )=(\left {14} \atop {5}} \right. )=2002$

iii. Assuming that the 5 pieces of chocolate are of different types and she isn't willing to give any friend more than one piece of chocolate, the situation is said to be WITHOUT REPLACEMENT & ORDERED

n = 10 & k = 5

Total number of possible combinations becomes,

$P\left {n} \atop {k}} \right=P\left {10} \atop {5}} \right. =30240$

iv. Assuming the the 5 pieces of chocolate are of different types, if she is willing to give a friend more than one piece of chocolate, the situation is said to be WITH REPLACEMENT & ORDERED

n = 10 AND k = 5

Total number of combinations becomes

$n^k=10^5=100,000$

Sampling with replacement means that one friend can be sampled more than once i.e: A friend receives a piece of chocolate more than once

The order dictates how the sampling is applied.

3 0

3 years ago

Draw a picture of a figure that has more than three lines of symmetry draw the lines of symmetry

Minchanka [31]

Draw a circle, then take and draw lines across it for however many you want. circles have infinite lines of symmetry so anything works.

6 0

3 years ago

. 1. A consistent system of equations is a system with __________.. . the same line. . parallel lines. . intersecting lines and

madreJ [45]

A consistent system of equations is a system with intersecting lines and lines that have the same equation.

A consistent system is the system of equations with at least one solution or infinite number of solutions.

So , the intersecting lines will always have a solution and the lines with the same equation will have infinite number of solution.

The graph of a system of equations with different slopes will have no solutions. Correct option in "Never"

If the system of equations have different slopes then the equations are not parallel, hence the equations will always have a solution.

3 0

3 years ago