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Andrews [41]
2 years ago
6

WILL MARK BRANLIEST IF GOTTEN RIGHT

Mathematics
1 answer:
Mamont248 [21]2 years ago
3 0

Answer:

The last choice. The red one

Step-by-step explanation:

I'm just guessing at this point. But, by the looks of it, red does seem like it would be correct. Again, this is just a guess. So, if it's wrong I'm so sorry. Have a lovely day.

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Sally is selling two types of seashells by the seashore. She is selling large and small shells. She sells small shells for 2 dol
Ostrovityanka [42]

Answer:

The answer is A

Step-by-step explanation:

If you plug in the numbers it looks like this:

(A)  5(2) + 10(5) = 60

(B)  10(2) + 5(5) = 45

(C) 7(2) + 8(5) = 54

(D)  8(2) + 7(5)

so a is the only one that equals 60

4 0
3 years ago
which equation shows the point slope form of the line that passes through (3,2) and has a slope of 1/3
gavmur [86]

Answer:

y=1/3x+1

Step-by-step explanation:

y=mx+b

y=1/3x+b

2=1/3(3)+b

2=1+b

1=b

y=1/3x+1

8 0
2 years ago
I need helppp!!! <br> −50=10s
Nimfa-mama [501]
Hey the answer is -0.2. You have to divide 10 by both sides to isolate the variable
4 0
2 years ago
Slope =1 /2 point passes through the (-8,1) what is the equation
Archy [21]

Hope it helps you...

Thank you!!

7 0
2 years ago
Suppose you choose a team of two people from a group of n &gt; 1 people, and your opponent does the same (your choices are allow
jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

 \left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right     i^{3}

Step-by-step explanation:

selecting the first team from n people we have \left(\begin{array}{ccc}n\\1\\\end{array}\right)  = n possibility and choosing second team from the rest of n-1 people we have \left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1

As { A, B} = {B , A}

Therefore, the total possibility is \frac{n(n-1)}{2}

Since our choices are allowed to overlap, the second team is \frac{n(n-1)}{2}

Possibility of choosing both teams will be

\frac{n(n-1)}{2}  *  \frac{n(n-1)}{2}  \\\\= [\frac{n(n-1)}{2}] ^{2}

We now have the formula

1³ + 2³ + ........... + n³ =[\frac{n(n+1)}{2}] ^{2}

1³ + 2³ + ............ + (n-1)³ = [x^{2} \frac{n(n-1)}{2}] ^{2}

=\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] =   [\frac{n(n-1)}{2}]^{3}

4 0
3 years ago
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