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2 years ago

Please help me with my math homework what is the answer!!!

Mathematics

1 answer:

Dmitry [639]2 years ago

6 0

Answer:

Step-by-step explanation:

The letters are easy enough.

y^3 determines that 3 ys are needed. y^2 is taken in by the cubed.

x^4 determines that 4 xs are needed. x^3 is part of x^4 and you don't need any more xs

12 and 42 are the parts that will cause the problem. Factor them both into prime factors

12: 2 * 2 * 3

42: 2 * 3 * 7

You need two 2s.

You need one 3

You need one 7

LCD = 2 *2*3 * 7 = 84

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What is the scale factor of the dilation of triangle DEF 3/10 3/7 7/3 7

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3 years ago

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What is the best first step for solving the given system using substitution while avoiding fractions? =9x+4x=10 -9x+3y=3

aivan3 [116]

Answer:

Solve for y in the second equation

Step-by-step explanation:

We assume your system is ...

-9x +4y = 10
-9x +3y = 3

Dividing the second equation by 3 gives ...

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so an expression for y without fractions can be found by solving for y, that is, by adding 3x:

y = 3x +1

_____

Comment on alternate solution

We'd be tempted to solve the first equation for -9x and substitute for that.

-9x = 10 -4y

(10 -4y) +3y = 3 . . . . . substitute for -9x

-y = -7 . . . . . . . . . . . . . simplify, subtract 10

y = 7 . . . . . . . . . . . . . . multiply by -1

3 0

3 years ago

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Dustin has a bag of marbles with 4 blue marbles, 4 white marbles, and 3 red marbles.

diamong [38]

Answer:

$\dfrac{6}{55}$

$\dfrac{4}{165}$

Step-by-step explanation:

Dustin has a bag of marbles with 4 blue marbles, 4 white marbles, and 3 red marbles. There are 11 marbles in total.

A. The probability that the first marble drawn is blue marble is

$\dfrac{4}{11},$

the probability that the second marble drawn is red marble is

$\dfrac{3}{10}$

The probability that the first marble drawn is blue and the second is red is

$\dfrac{4}{11}\cdot \dfrac{3}{10}=\dfrac{12}{110}=\dfrac{6}{55}$

B. The probability that the first marble drawn is red marble is

$\dfrac{3}{11},$

the probability that the second marble drawn is white marble is

$\dfrac{4}{10}$

The probability that the first marble drawn is red and the second is white is

$\dfrac{3}{11}\cdot \dfrac{4}{10}=\dfrac{12}{110}=\dfrac{6}{55}$

C. The probability that the first marble drawn is blue marble is

$\dfrac{4}{11},$

the probability that the second marble drawn is blue marble is

$\dfrac{3}{10},$

the probability that the third marble drawn is blue marble is

$\dfrac{2}{9},$

The probability that the first, second and third marbles drawn are blue is

$\dfrac{4}{11}\cdot \dfrac{3}{10}\cdot \dfrac{2}{9}=\dfrac{24}{990}=\dfrac{4}{165}$

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3 years ago