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3 years ago

Write 5(6x+4)-2(5x-2) in the form a(bx+c) where a,b and c integers and a>1

Mathematics

1 answer:

Tamiku [17]3 years ago

6 0

Answer:

Step-by-step explanation:

$5(6x+4)-2(5x-2)\qquad\text{use distributive property}\ a(b+c)=ab+ac\\\\=(5)(6x)+(5)(4)+(-2)(5x)+(-2)(-2)\\\\=30x+20-10x+4\qquad\text{combine like terms}\\\\=(30x-10x)+(20+4)\\\\=20x+24$

$20x=4\cdot5x\\\\24=4\cdot6\\\\20x+24=4(5x+6)$

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Cx = r + d solve for x

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Answer:

x = (r+d)/c

If all you need is for the x to stand alone, all you must do is get rid of all of the values on its side of its equation. So in the equation cx=r+d, the only value with x is c. To get rid of this, you divide it on both sides of the equation to get x = (r+d)/c.

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3 years ago

You put $5,000 in a simple interest account at a bank. You will earn $1,200 in just 4 years. What is the annual interest rate fo

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The annual interest rate is 6%

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3 years ago

A baseball coach needs to choose 99 players to be in the batting lineup for the next game. There are 44 freshmen, 44 sophomores,

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Answer:

The number of ways the batting line up can be made is 3,851,971,200 ways.

Step-by-step explanation:

The coach of the baseball team needs to choose 9 players for the batting lineup.

He can choose from a total of 4 freshmen, 4 sophomores, 5 juniors and 4 seniors for the team.

He has to choose the team so that there at most 2 juniors on the lineup.

The total number of available player who are not juniors is 4 + 4 + 4 = 12

i) If the coach selects no juniors then there are 9 positions to fill up and

since the batting order is not specific so these 9 players can be selected

from 12 players = $\binom{12}{9}$ = 220 ways

ii) If the coach selects one junior then there are other 8 other positions to

fill up from 12 players and one junior is to be selected from 5 juniors

= $\binom{12}{8} \times \binom{5}{1} = 2475$

iii) If the coach selects two juniors then there are other 7 other positions to

fill up from 12 players and two juniors are to be selected from 5 juniors

= $\binom{12}{7} \times \binom{5}{2} = 7920$

So the ways of selecting the players is = 220 + 2475 + 7920 = 10615 ways

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3 years ago

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Answer:

landing on a shaded portion and landing on an even number ; landing on a shaded portion and landing on a number greater than 3 ; landing on an unshaded portion and landing on an odd number ; and landing on an unshaded portion and landing on a number less than 2.

Explanation:

Mutually exclusive events are events that cannot occur at the same time. Our spinner has two grey sections, on 1 and 3; and two white sections, on 2 and 4.

This means that the spinner cannot land on a shaded (grey) portion and land on an even number at the same time, since the grey sections are numbered 2 and 4, both of which are even numbers.

The spinner also cannot land on a shaded (grey) portion and land on a number greater than 3 at the same time; this is because the only number greater than 3 on the spinner is 4, which is a white portion.

The spinner cannot land on an unshaded (white) portion and land on an odd number, since the white sections are labeled 2 and 4, which are both even.

The only number on the spinner less than 2 is 1, which is grey; this means the spinner cannot land on a number less than 2 and an unshaded (white) portion at the same time.

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4 years ago

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Answer:

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