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RUDIKE [14]
3 years ago
8

B) Is the relation a function? Use the vertical line test to decide.

Mathematics
1 answer:
Sedaia [141]3 years ago
8 0

Answer:

Step-by-step explanation:

The answer is A U just did that on Khan academy

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Change the subject of the equation v = u + at to a.
emmasim [6.3K]
<h3>Answer ↓</h3>

<h3>Calculations ↓</h3>

In order to make a the subject of this equation , we need to get a by itself .

The current equation is :

v = u + at

Subtract u on both sides :

v-u=at

Now, divide by t on both sides :

v-u/t=a

<h3>So the formula looks like ↓</h3>

\boxed{\\\begin{minipage}{3cm}Equation \\ a=$\displaystyle\frac{v-u}{t} $ \\ \end{minipage}}

hope helpful ~

8 0
2 years ago
In the figure below, if angle ZYX measures 45 degrees, then arc XY measures 45 degrees.
Daniel [21]
I think the answer is true.
4 0
3 years ago
The function r(n) = 5n² + 1 represents the value of the nth term in a sequence.
umka21 [38]

Step-by-step explanation:

a_n=5n²+1

➜a_1=5(1)²+1

➜5×1+1

➜5+1

<h3>➜6</h3>

➜a_3=5(3)²+1

➜5×9+1

➜45+1

<h3>➜46</h3>
7 0
3 years ago
What’s 2+2= I need it just cuss
MArishka [77]

Answer:

2+2 is 4 1+1+1+1 is 4 1234

5 0
4 years ago
Read 2 more answers
* 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

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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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