Please answer this for me correctly. There are two questions. THANK YOU!!

9x^2-1-(3x-2)^2=0 find the roots

Marizza181 [45]

Step-by-step explanation:

Make me as brain liest plz i need it

3 0

3 years ago

HELP PLEASE 40 POINTS The graph shows the prices of different numbers of bushels of corn at a store in the current year. The tab

Vesna [10]

Answer:

see below

Step-by-step explanation:

Part A

Since the lines goes through the point (0,0) the graph is proportional. We can find the rate of change by take the price of corn and dividing by the number of bushels

24/3 = 8 dollars/ bushel

Part B

Previous Year Number of Bushels Price of Corn (dollars)

3 21

6 42

9 63

12 84

We can find the rate of change for the previous year by using the slope formula

m = (y2-y1)/(x2-x1)

m = (84-63)/(12-9)

=21 / 3

= 7

The previous year was 7 dollars per bushel

The increase was 8-7 = 1 dollar per bushel

4 0

3 years ago

Read 2 more answers

What is the value of x? 6x−2(x 4)=12

jonny [76]

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step 2 : 6x2 - x - 4 = 0 ((2•3x2) - x) - 4 = 0 Step 2 :Trying to factor by splitting the middle term

2.1 Factoring 6x2-x-4

The first term is, 6x2 its coefficient is 6 .
The middle term is, -x its coefficient is -1 .
The last term, "the constant", is -4 

Step-1 : Multiply the coefficient of the first term by the constant 6 • -4 = -24

Step-2 : Find two factors of -24 whose sum equals the coefficient of the middle term, which is -1 .

-24 + 1 = -23 -12 + 2 = -10 -8 + 3 = -5 -6 + 4 = -2 -4 + 6 = 2 -3 + 8 = 5 -2 + 12 = 10 -1 + 24 = 23

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step 2 : 6x2 - x - 4 = 0

3 0

3 years ago

Fill in the blanks below using the word bank.

Charra [1.4K]

M is the slope.

B is the y-intercept

The constant of proportionality is the ratio between two directly proportional quantities.

7 0

3 years ago

Read 2 more answers

Use series to verify that <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago