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3 years ago

Answer correctly for brainleist and 5 mark

Mathematics

1 answer:

notsponge [240]3 years ago

8 0

Answer:

It is the last answer.

Step-by-step explanation:

If you graph this, there will be no point more negative than (-4/3, -5).

This means the x and y values must be greater than or equal to these values, as the fourth answer will show.

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I really dislike trigonometry, here's another question

alex41 [277]

It’s a little blurry but i can help

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3 years ago

2. A record store is going out of business and sold any

liubo4ka [24]

Answer:

A) c + CD = 20

B) 6c + 12CD = 192

Multiplying equation A by -6 we get:

A) -6c -6CD = -120

B) 6c + 12 CD = 192

Adding the equations we get

0 + 6 CD = 72

There are 12 CD's and

c + 12 = 20

There are 8 cassettes

Step-by-step explanation:

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2 years ago

All expression equal to 4x4x4

rusak2 [61]

Answer:

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Step-by-step explanation:

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3 years ago

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Given APWVS ~ HDTFY, what is HD?

SIZIF [17.4K]

The answer is B:15
since APWVS ~ HDTFY, PW ~ DT. so you set up a proportion. i hope this explains it clearly for you, if not i'm sorry

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3 years ago

* Some amount of billiard balls were arranged in an equilateral triangle. And 5 balls were extra. When the same set of billiard

Agata [3.3K]

Lets say billiard balls are arranged in rows to form an equilateral triangle, then the first row consists of 1 ball, second row consists of 2 balls, and third row consists of 3 balls, and so on. So there must be $n$ balls in the $n^{th}$ row.

So, the total number of balls that forms the equilateral triangle with $n$ rows is:

$1+2+3+4+5+....+n=\frac{n(n+1)}{2}$

Let $x_1$ and $x_2$ be the total number of balls in the first and second arrangements respectively.

Then,

$x_1=\frac{n(n+1)}{2} +5$

It has been said that there were 11 lesser balls in the second arrangement:

$x_2=\frac{1+(n+1)}{2} \times (n+1)-11=(n+1) \times \frac{(n+2)}{2} -11$

Since, $x_1=x_2$

$\frac{(n+1)}{2} \times n+5=\frac{(n+2)}{2} \times(n+1)-11$

multiplying both the sides by 2

$(n+1)\times n+10=(n+2)(n+1)-22$

$n+n^2=n^2+n+2n+2-22-10$

$2n=22+10-2$

$2n=30$

$n=15$

Therefore,

$x_1=\frac{(n+1)}{2}\times n+5=\frac{15+1}{2} \times 15+5=125$

So, there were 125 balls at the set.

4 0

3 years ago

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