Which list contains three points that lie on the graph of the function?

I really need help Can someone help me

denis-greek [22]

The answer is 3.800 because you round up

6 0

3 years ago

Someone who knows the remainder theorem help!

Damm [24]

Answer:

○ D. Yes, x = 12 is a zero of the polynomial.

The quotient is x + 22, and the remainder is 0.

Step-by-step explanation:

On a second thought, I knew something similar to that theorem because factoring them would determine if it has a remainder:

[x - 12][x + 22]

I am joyous to assist you anytime.

* I apologize for the previous answer I gave you.

5 0

4 years ago

You know that 4 pizzas will feed 16 people how many pizzas do you need to feed 88 people?

Furkat [3]

You can multiply 16 times what equls 88 and then divide by 4 much more simple than it sounds and also your in high school and u cant solve this problem

6 0

3 years ago

30 points!!<br> What is the sum of the first six terms of the series?<br> 48 - 12 + 3 - 0.75 +...

Lunna [17]

Answer:

The sum of the first six terms is 38.39

Step-by-step explanation:

This is a geometric sequence since the common difference between each term is $-\frac{1}{4}$

Thus, $r=-\frac{1}{4}$

To find the sum of first six terms, we need to find the fifth and sixth term of the sequence.

To find the fifth term:

The general form of geometric sequence is $a_{n}=a_{1} \cdot r^{n-1}$

To find the fifth term, substitute $n=5$ in $a_{n}=a_{1} \cdot r^{n-1}$

$\begin{aligned}a_{5} &=(48) \cdot\left(-\frac{1}{4}\right)^{5-1} \\&=(48) \cdot\left(-\frac{1}{4}\right)^{4} \\&=(48)\left(\frac{1}{256}\right) \\a_{5} &=0.1875\end{aligned}$

To find the sixth term, substitute $n=6$ in $a_{n}=a_{1} \cdot r^{n-1}$

$\begin{aligned}a_{6} &=(48) \cdot\left(-\frac{1}{4}\right)^{6-1} \\&=(48) \cdot\left(-\frac{1}{4}\right)^{5} \\&=(48)\left(-\frac{1}{1024}\right) \\a_{5} &=-0.046875\end{aligned}$

To find the sum of the first six terms:

The general formula to find Sn for $|r|$ is $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$

$\begin{aligned}S_{6} &=\frac{48\left(1-\left(-\frac{1}{4}\right)^{6}\right)}{1-\left(-\frac{1}{4}\right)} \\&=\frac{48\left(1-\frac{1}{4096}\right)}{1+\frac{1}{4096}} \\&=\frac{48(0.95)}{5} \\&=\frac{48(0.9998)}{5} \\&=\frac{48(0.9998)}{5} \\&=\frac{47.9904}{5} \\&=38.39\end{aligned}$