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3 years ago

Which factors can be multiplied together to make the trinomial 5x2 + 8x – 4? Check all that apply.

Mathematics

2 answers:

Mrrafil [7]3 years ago

3 0

Answer:

x+2

5x-2

Step-by-step explanation:

There is only 2 binomials that will multiply together that will give the given trinomial.

ax^2+bx+c

5x^2+8x-4

Goal: Find two numbers that multiply to be a*c and add up to be b.

We will use this goal to factor our trinomial.

a=5

b=8

c=-4

--------

a*c=-20

b=8

Can you think of two numbers that multiply to be -20 and add up to be 8? I hope you said 10 and -2.

So we are going to replace 8x with -2x+10x.

5x^2+8x-4

5x^2-2x+10x-4

We are going to pair the first terms together and the second two terms together like so:

(5x^2-2x)+(10x-4)

Now we are going to factor each pair.

x(5x-2)+2(5x-2)

(5x-2)(x+2)

So (x+2) is a factor of the given trinomial and (5x-2) is a factor of the given trinomial.

faust18 [17]3 years ago

3 0

Answer

x+2 and 5x-2

Step-by-step explanation:

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The cost of a book was decreased from 60 to 50 by what percent the price decrased with solution

Artist 52 [7]

Answer:

The price decreased by 16 1/3%, or approximately 16.3%

Step-by-step explanation:

Use the formula for percent change. If the answer is negative, it is a percent decrease. if the answer is positive, it is a percent increase.

percent change = (new price - old price)/(old price) * 100%

In this case, use:

new price = 50

old price = 60

percent change = (new price - old price)/(old price) * 100%

percent change = (50 - 60)/60 * 100%

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3 years ago

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3 years ago

2,17,82,257,626,1297 next one please ?

In-s [12.5K]

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule $n^4+1$ . The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the n-th term in this sequence by $a_n$ , and denote the given sequence by $\{a_n\}_{n\ge1}$ .

Let $b_n$ denote the n-th term in the sequence of forward differences of $\{a_n\}$ , defined by

$b_n=a_{n+1}-a_n$

for n ≥ 1. That is, $\{b_n\}$ is the sequence with

$b_1=a_2-a_1=17-2=15$

$b_2=a_3-a_2=82-17=65$

$b_3=a_4-a_3=175$

$b_4=a_5-a_4=369$

$b_5=a_6-a_5=671$

and so on.

Next, let $c_n$ denote the n-th term of the differences of $\{b_n\}$ , i.e. for n ≥ 1,

$c_n=b_{n+1}-b_n$

so that

$c_1=b_2-b_1=65-15=50$

$c_2=110$

$c_3=194$

$c_4=302$

etc.

Again: let $d_n$ denote the n-th difference of $\{c_n\}$ :

$d_n=c_{n+1}-c_n$

$d_1=c_2-c_1=60$

$d_2=84$

$d_3=108$

etc.

One more time: let $e_n$ denote the n-th difference of $\{d_n\}$ :

$e_n=d_{n+1}-d_n$

$e_1=d_2-d_1=24$

$e_2=24$

etc.

The fact that these last differences are constant is a good sign that $e_n=24$ for all n ≥ 1. Assuming this, we would see that $\{d_n\}$ is an arithmetic sequence given recursively by

$\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}$

and we can easily find the explicit rule:

$d_2=d_1+24$

$d_3=d_2+24=d_1+24\cdot2$

$d_4=d_3+24=d_1+24\cdot3$

and so on, up to

$d_n=d_1+24(n-1)$

$d_n=24n+36$

Use the same strategy to find a closed form for $\{c_n\}$ , then for $\{b_n\}$ , and finally $\{a_n\}$ .

$\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}$

$c_2=c_1+24\cdot1+36$

$c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2$

$c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3$

and so on, up to

$c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)$

Recall the formula for the sum of consecutive integers:

$1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2$

$\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)$

$\implies c_n=12n^2+24n+14$

$\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}$

$b_2=b_1+12\cdot1^2+24\cdot1+14$

$b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2$

$b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3$

and so on, up to

$b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)$

Recall the formula for the sum of squares of consecutive integers:

$1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$

$\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)$

$\implies b_n=4n^3+6n^2+4n+1$

$\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}$

$a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1$

$a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2$

$a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3$

$\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1$

$\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4$

$\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)$

$\implies a_n=n^4+1$

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3 years ago

Tim answered all the questions on his math test but got 10 answers wrong. He received 4 points for every correct answer, and the

charle [14.2K]

4x = 76
Where x is how many awnsers he got correct and 4 how many points each correct question awnsered.

Now you just have to isolate x by dividing both sides of the equation by 4

4x/4 = 76/4
x = 19

So he got 19 questions right and 10 awnsers wrong. In total there were 29 questions on his test

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4 years ago

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