3x^2+ 4x + 5 for x = -2

Mathematics

1 answer:

Alexus [3.1K]3 years ago

7 0

The answer is 33
3(-2)^2+4(-2)+5
-6^2+ (-8) + 5
36 + (-8) + 5
28+5
33

You might be interested in

Please help me with these two math problems that I do not quite understand. This is urgent!

g100num [7]

Answer:

1) $a_{n}=a_{1}+(n-1)d\Rightarrow a_{n}=-1-2(n-1)\\a_{2}=-1-2(2-1)\Rightarrow a_{2}=-3\\a_{3}=-1-2(3-1)\Rightarrow a_{3}=-5\\(...)\\a_{10}=-1-2(10-1)=-19$

2) $a_{10}=4+8(10-1)\Rightarrow a_{10}=76$

Step-by-step explanation:

1) To write an Arithmetic Sequence, as an Explicit Term, is to write a general formula to find any term for this sequence following this pattern:

$a_{n}=a_{1}+d(n-1)\Rightarrow \left\{\begin{matrix}a_{n}=n^{th}\: term\\a_1=1st \: term \\d=\: difference\\n=n^{th}\, term\end{matrix}\right.$

<em>"Write an explicit formula for each explicit formula A(n)=-1+(n-1)(-2)"</em>

This isn't quite clear. So, assuming you meant

Write an explicit formula for each term of this sequence A(n)=-1+(n-1)(-2)

As this A(n)=-1+(n-1)(-2) is already an Explicit Formula, since it is given the first term $a_{1}=-1$ the common difference $d=-2$ let's find some terms of this Sequence through this Explicit Formula:

$a_{n}=a_{1}+(n-1)d\Rightarrow a_{n}=-1-2(n-1)\\a_{2}=-1-2(2-1)\Rightarrow a_{2}=-3\\a_{3}=-1-2(3-1)\Rightarrow a_{3}=-5\\(...)\\a_{10}=-1-2(10-1)=-19$

2) $(4,12,20,28, ..)$ In this Arithmetic Sequence the common difference is 8, the first term value is 4.

Then, just plug in the first term and the common difference into the explicit formula:

$a_{10}=4+8(10-1)\Rightarrow a_{10}=76$

6 0

3 years ago

(07.01)

Zinaida [17]

The answer is $3,045.48

6 0

3 years ago

Read 2 more answers

Find the median, lower quartile, and upper quartile for the set of data.

harkovskaia [24]

$\text{Find the median:}\\\\\text{To find the median, you would cross out the numbers from the side}\\\text{until you're left with the remainders. Put them in order and solve}\\\\1, 2, 4 ,5, 6, 7, 8, 9, 10, 11\\\text{Add 6 and 7 then divide by 2}\\\\6+7=13\\\\11\div2=6.5\\\\\boxed{\text{The median is 6.5}\\}\\\\\text{The lower quartile is the median of the numbers less than the median, so}\\\text{the numbers from 1-6}\\\\\text{When you cross out the numbers, you would be left with 4}$

$\boxed{\text{The lower quartile is 4}}\\\\\text{Do the same for the upper quartile for the numbers greater than half}\\\text{of the data set}\\\\\boxed{\text{Your upper quartile would be 9}}$