All categories

3 years ago

What is 27/33 is the lowest term?

Mathematics

2 answers:

MariettaO [177]3 years ago

6 0

9/11

Hope this helps

belka [17]3 years ago

4 0

$\frac{27}{33} = \frac{3*9}{3*11} = \frac{9}{11}$

In short, the answer would be : $\frac{9}{11}$ .

Hope this helps !

Photon

You might be interested in

Emily is entering a bicycle race for charity. Her mother pledges $0.90 for every 0.25 mile she bikes. If Emily bikes 7 miles, ho

mestny [16]

Answer:

Find how much she will be paid per mile to make it easier.

.25 * 4 = 1

.90 * 4 = 3.60

Now you know she earns 3.60 per mile, multiply that by 7.

3.60 * 7 = $25.20 total.

6 0

3 years ago

Read 2 more answers

Formulate the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- an

slega [8]

Answer:

1) There were 33 $4,000 investors and 27 $8,000 investors.

2) The solution in x = 4, y = 9.

3) There were 24 nickels and 56 dimes.

Step-by-step explanation:

1) A lawyer has found 60 investors for a limited partnership to purchase an inner-city apartment building, with each contributing either $4,000 or $8,000. If the partnership raised $348,000, then how many investors contributed $4,000 and how many contributed $8,000?

I am going to say that:

x is the number of investors that contributed 4,000.

y is the number of investors that contributed 8,000.

Building the system:

There are 60 investors. So:

$x + y = 60$

In all, the partnership raised $348,000. So:

$4000x + 8000y = 348000$

I am going to simplify by 4000. So:

$x + 2y = 87$

Solving the system:

The elimination method is a method in which we can transform the system such that one variable can be canceled by addition. So:

$1)x + y = 60$

$2)x + 2y = 87$

I am going to multiply 1) by -1. So we have

$1)-x - y = -60$

$2)x + 2y = 87$

By addition, the x are going to cancel each other

$-x + x - y + 2y = -60 + 87$

$y = 27$

For x:

$x + y = 60$

$x = 60-y = 60-27 = 33$

There were 33 $4,000 investors and 27 $8,000 investors.

2) Solve the system by row-reducing the corresponding augmented matrix.

$2x + y = 17$

$x + y = 13$

This system has the following augmented matrix:

$\left[\begin{array}{ccc}2&1&17\\1&1&13\end{array}\right]$

To help the row reducing, i am going to swap the first with the second line:

$L1 L2$

So we have:

$\left[\begin{array}{ccc}1&1&13\\2&1&17\end{array}\right]$

Now, reducing the first column.

$L2 = L2 - 2L1$

So we have:

$\left[\begin{array}{ccc}1&1&13\\0&-1&-9\end{array}\right]$

Now we do:

$L2 = -L2$

And the matrix is:

$\left[\begin{array}{ccc}1&1&13\\0&1&9\end{array}\right]$

Now to reduce the second column, we do:

$L1 = L1 - L2$

$\left[\begin{array}{ccc}1&0&4\\0&1&9\end{array}\right]$ .

So the solution is:

x = 4, y = 9.

3) A jar contains 80 nickels and dimes worth $6.80. How many of each kind of coin are in the jar?

I am going to say that x is the number of nickels and y is the number of dimes.

Each nickel is worth 5 cents and each dime is worth 10 cents.

Building the system:

There are 80 coins in all:

$x + y = 80$

They are worth $6.80. So:

$0.05x + 0.10y = 6.80$

Solving the system:

$1)x + y = 80$

$2)0.05x + 0.10y = 6.80$

I am going to divide 1) by -10, so we can cancel y. So:

$1)-0.10x - 0.10y = -8$

$2)0.05x + 0.10y = 6.80$

Adding:

$-0.10x + 0.05x - 0.10y + 0.10y = -8 + 6.80$

$-0.05x = -1.2$ *(-100)

$5x = 120$

$x = \frac{120}{5}$

$x = 24$

Also

$x + y = 80$

$y = 80-x = 80-24 = 56$

There were 24 nickels and 56 dimes.

8 0

3 years ago

Ms. Spencer is packing orders for her bath supply business. One customer ordered 8 bath fizzies and 4 jars of sugar scrub. Each

sergey [27]

8 bath fizzes * 5 (ounces) = 40 ounces. 4 jars of sugar scrub * 1.5 (pounds) = 6 pounds. So 6 pound and 40 ounces. (40 ounces = 2.5 pounds. So 8.5 pounds in total.) ( If you want to convert the answer to ounces, 8.5 pounds is 136 ounces.)

8 0

3 years ago

Can you help me with this please !!! Please really please

Stolb23 [73]

Answer:

brainliest pls???

Step-by-step explanation:

8 0

4 years ago

Find the length of the shorter diagonal of a rhombus if:

Fudgin [204]

a) Area of rhombus = ½ * (diagonal1)*(diagonal 2)