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3 years ago

I’m really confused could someone help me?

Mathematics

2 answers:

Vilka [71]3 years ago

8 0

Answer:

The answer to your question is: a and b are solutions,

96 crackers

Step-by-step explanation:

a) (4, 0)

y = x - 4

0 = 4 -4

0 = 0 It's a solution

b) (3, -1)

y = x - 4

-1 = 3 - 4

-1 = -1 It's a solution

c) (6, 3)

y = x - 4

3 = 6 - 4

3 = 2 It's not a solution

d) (2, -4)

y = x - 4

-4 = 2 - 4

-4 = -2 It's not a solution

Second question, 96 crackers

8 crackers ------------------- 1 serving

x --------------------- 12 serving

x = (12 x 8) / 1

x = 96 crackers

GenaCL600 [577]3 years ago

6 0

Answer:

B and C

Step-by-step explanation:

4,0 x=4 y=x-4=0

3,-1 x=4 y=x-4=-1

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50 is 25% of what number

Marat540 [252]

200. 50x4=200 hope that helps

5 0

3 years ago

Read 2 more answers

What is the missing number in the sequence below? 39, 24, 9, ____, -21

Oksi-84 [34.3K]

Answer:

-6

Step-by-step explanation:

The pattern here is that you substract 15 every time you move a number (left to right) , therefore: 9-15 = -6

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below =)

6 0

3 years ago

Suppose you choose a team of two people from a group of n > 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

4 years ago

Find FC A. 10.6 B. 12.6 C. 14 D. 180 Helppppppp ASAP please

Cloud [144]

The answer is 12.6 for sure

6 0

3 years ago

Match the provided functions to a graphed function with the same zero(s).

satela [25.4K]

Answer:

1st:

${x}^{3} - 3 { x}^{2} - 3$

2nd:

${x}^{2} - 5x + 4$

3rd:

$- 4x - 8$

Step-by-step explanation:

Graph all of them and see which ones cross the x- axis at the same points.

8 0

3 years ago