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2 years ago

Find the greatest common factor of 15x^6 and 33x^4y^5

Mathematics

2 answers:

Katena32 [7]2 years ago

5 0

Answer:

495x^10 y^5

Step-by-step explanation:

zhannawk [14.2K]2 years ago

4 0

Answer:

3x^4

Step-by-step explanation:

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( ONLY ANSWER THIS IF YOU KNOW WHAT THE ANSWER IS) I will give Brainliest!

vodka [1.7K]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

SUBSTITUTING INGREDIENTS

tatiyna

Answer:

1. 1 tablespoon Dijon mustard

2. Combine six Tablespoons of cocoa powder and two Tablespoon of vegetable oil, butter or shortening

3. 2 cups of all-purpose flour, 3 teaspoons baking powder, and ½ teaspoon salt

3. One teaspoon of dried basil leaves

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Step-by-step explanation:

CAn I have brainliest? TYSMMMMMMMMMM

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3 years ago

Solve the system of linear equations. separate the x- and y- values with a coma. -5x+6y=33 -18x+4y=-22

Pavel [41]

I hope I can help you:)

4 0

1 year ago

Mathswatch Question! Mr Powers checks his electricity bill. Here are his meter readings. New reading: 6538 units, Old reading: 6

Setler79 [48]

Answer:

$Cost = 284 x$

Step-by-step explanation:

Given

$Old = 6245\ units$

$New = 6538\ units$

Required

The cost of units used

First, calculate the number of units used.

$Units = New - Old$

$Units = 6538 - 6254$

$Units = 284$

Next, multiply the units used by the cost per units,

Since the cost per unit is not given, we assume that it is x.

So, the total cost is:

$Cost = Units * Cost\ per\ units$

$Cost = 284 * x$

$Cost = 284 x$

3 0

3 years ago

If a = √3-√11 and b = 1 /a, then find a² - b²

Serga [27]

If $b=\frac1a$ , then by rationalizing the denominator we can rewrite

$b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8$

Now,

$a^2 - b^2 = (a-b) (a+b)$

and

$a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8$

$a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8$

$\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}$

5 0

1 year ago