Can u plz answer this question

Quick anyone know the answer to this problem? Will give brainiest

natka813 [3]

Answer:

I think its b.

5 0

3 years ago

Read 2 more answers

What is the rate of change of the linear function represented by the table? StartFraction one Over three EndFraction StartFracti

Lynna [10]

The rate of change of the linear function represented by the table is 3

<h3>Rate of change of a function</h3>

The rate of change of a function is also known as the slope of a function. The equation for calculating the slope is expressed as:

Slope = y2-y1/x2-x1

using the coordinates from the table (-2, -13) and (-1, -10)

Substitute

Slope = -10-(-13)/-1-(-2)

Slope = -10+13/-1+2

Slope =3

Hence the rate of change of the linear function represented by the table is 3

Learn more on rate of change here: brainly.com/question/25184007

#SPJ1

5 0

2 years ago

(7x²+2x+5)+(3x²-8x-10)

Mariulka [41]

First you would simplify the equation <span>(7x²+2x+5)+(3x²-8x-10) lets start with the first part
</span><span>(7x²+2x+5)
</span>7x² + 2x which = 49x + 2x which also = 51x Then you add the last part (+5)
51x + 5
Now you simplify the last part.
<span>(3x²-8x-10)
</span>3x²-8x which=9x - 8x which also = 1x or simply x Then add the last part (-10) Which you'll get x-10
51x + 5) + (x - 10) = 52x -5
so your answer is 52x - 5

7 0

3 years ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago

What is the solution of 3+x-2/x-3<_4

VikaD [51]

Step-by-step explanation:

3+x-2/x-3<_4

cross multiply

3+x-2<_4(x-3)

3+x-2<_4x-12

1+x<_4x-12

collect like terms

1+12<_4x-x

13<_3x

divide both side by 3

13/3<_×

6.5<_x

3 0

3 years ago