I need help, please, I really need a help, please.

A part of Jennifer’s work is to solve the equation 2(6x^2-3)=11x^2-x is shown below

kotykmax [81]

Answer:

Step-by-step explanation:

2(6x² - 3) = 11x² - x

2*6x² - 2*3 = 11x² - x

12x² - 6 = 11x² -x

Subtract 11x² from both sides

12x² - 11x² - 6 = -x

x² - 6 = -x

x² + x - 6= 0

Sum =1

Product = -6

Factors = 3 , (-2) { 3*(-2) = -6 & 3 +(-2) = 1}

x² + 3x - 2x - 6 = 0

x(x + 3) - 2(x + 3)= 0

(x +3)(x - 2) = 0

7 0

3 years ago

What is the vertex of the absolute value function below?

Nonamiya [84]

Answer:

sry I just wanted the points I'm in middle school so I don't know this stuff either but can you give free brainlyest I'm soo close to my next rank I'd really appreciate it if you would

4 0

2 years ago

I need help on 19 and 20. Please help me. Please don’t put a link. Please help me.

lianna [129]

Answer:

19:

4 < third side < 14

20:

3 < third side < 13

Step-by-step explanation:

hint: go to http://www.17 28.org/trianinq.htm

(fix the space between 17 and 28)

you're welcome xx

6 0

3 years ago

Futhe Mathematics <img src="https://tex.z-dn.net/?f=%28Cos%20%7B%7D%5E%7B4%7Dt%20-Sin%20%7B%7D%5E%7B4%7Dt%20%29%20%20%5Cdiv%2

Nastasia [14]

Answer:

cos2t/cos²t

Step-by-step explanation:

Here the given trigonometric expression to us is ,

$\longrightarrow \dfrac{cos^4t - sin^4t }{cos^2t }$

We can write the numerator as ,

$\longrightarrow \dfrac{ (cos^2t)^2-(sin^2t)^2}{cos^2t }$

Recall the identity ,

$\longrightarrow (a-b)(a+b)=a^2-b^2$

Using this we have ,

$\longrightarrow \dfrac{(cos^2t + sin^2t)(cos^2t-sin^2t)}{cos^2t}$

Again , as we know that ,

$\longrightarrow sin^2\phi + cos^2\phi = 1$

Therefore we can rewrite it as ,

$\longrightarrow \dfrac{1(cos^2t - sin^2t)}{cos^2t}$

Again using the first identity mentioned above ,

$\longrightarrow \underline{\underline{\dfrac{(cost + sint )(cost - sint)}{cos^2t}}}$

Or else we can also write it using ,

$\longrightarrow cos2\phi = cos^2\phi - sin^2\phi$

Therefore ,

$\longrightarrow \underline{\underline{\dfrac{cos2t}{cos^2t}}}$

And we are done !

$\rule{200}{4}$

Additional info :-

Derivation of cos²x - sin²x = cos2x :-

We can rewrite cos 2x as ,

$\longrightarrow cos(x + x )$

As we know that ,

$\longrightarrow cos(y + z )= cosy.cosz - siny.sinz$

So that ,

$\longrightarrow cos(x+x) = cos(x).cos(x) - sin(x)sin(x)$

On simplifying,

$\longrightarrow cos(x+x) = cos^2x - sin^2x$

Hence,

$\longrightarrow\underline{\underline{cos (2x) = cos^2x - sin^2x }}$

$\rule{200}{4}$

7 0

2 years ago

What equation contains the given point ( -3,5)

Alenkasestr [34]

Answer:

y = x + 8

Step-by-step explanation:

6 0

3 years ago

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