All categories

2 years ago

Find the prime factors. 7) 48

Mathematics

1 answer:

FinnZ [79.3K]2 years ago

7 0

Step-by-step explanation:

$48 = 48 \times 1 \\ 48 = (24 \times 2) \times 1 \\ 48 = (12 \times 2) \times 2 \times 1 \\ 48 = (6 \times 2) \times 2 \times 2 \times 1 \\ 48 = (3 \times 2) \times 2 \times 2 \times 2 \times 1$

You might be interested in

Which expression is equivalent to (4x^(3)y^(5))(3x^(5)y)^(2)

marshall27 [118]

Answer:

$(4x^3y^5)(3x^5y)^2 = 36*x^{13}y^7$

Step-by-step explanation:

Given

$(4x^3y^5)(3x^5y)^2$

Required

The equivalent expression

We have:

$(4x^3y^5)(3x^5y)^2$

Expand

$(4x^3y^5)(3x^5y)^2 = 4x^3y^5*9x^{10}y^2$

Further expand

$(4x^3y^5)(3x^5y)^2 = 4*9*x^3*x^{10}y^5*y^2$

Apply laws of indices

$(4x^3y^5)(3x^5y)^2 = 36*x^{13}y^7$

8 0

3 years ago

Help on 3 I do not get it

Rama09 [41]

The answer is nine hope this helps :)

8 0

3 years ago

In kite WXYZ, the measure of x=z=86° and y=72° What is the measure of w?

Anna [14]

W=166 degrees

okay so the total measure of the angles should be 360 so you gotta do 86+86+72+w=360
86+86+72=244
360-244=116

8 0

3 years ago

Read 2 more answers

How many square feet of paper do you need to cover a cube with a side 10in long

JulsSmile [24]

10in = 0.833333ft^2 that is the total amount for one side. For a cube that has 6 sides, you multiply 0.833333 by 6.

You need 4.999998ft^2 of paper

4 0

3 years ago

a. Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability th

hodyreva [135]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

3 or less than 3 tires among the 10 tires should be blemished.

There are some possibilities.

1000 tires are blemished, (5000 - 1000) = 4000 tires are not blemished.

From the 5000 tires, 10 tires can be chosen in $^{5000}C_{10}$ ways.

Possibility 1: 3 tires are blemished.

Total 10 tires can be chosen with 3 blemished tires in $^{1000}C_3 \times {4000}C_7$ ways.

Possibility 2: 2 tires are blemished.

Total 10 tires can be chosen with 2 blemished tires in $^{1000}C_2 \times ^{4000}C_8$ ways.

Possibility 3: 1 tires are blemished.

Total 10 tires can be chosen with 1 blemished tires in $^{1000}C_1 \times^{4000}C_9$ ways.

Possibility 4: No tires are blemished.

Total 10 tires can be chosen with no blemished tires in $^{1000}C_0 \times^{4000}C_{10}$ ways.

Hence, the required probability is $\frac{^{1000}C_1 \times^{4000}C_9 + ^{1000}C_0 \times^{4000}C_{10} + ^{1000}C_2 \times^{4000}C_8 + ^{1000}C_3 \times^{4000}C_7}{^{5000}C_{10} }$ .

6 0

3 years ago