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3 years ago

2(x-1)+3x=3 Distributive property

Mathematics

2 answers:

RoseWind [281]3 years ago

8 0

X=1 its easy just distribute the 2 into it combine like terms and solve for x

Allisa [31]3 years ago

3 0

Answer:

x = 1

Step-by-step explanation:

Firstly we distribute: 2x - 2 + 3x = 3

Collect like terms: 5x - 2 + 3

Then add 2 to both sided to isolate 5x: 5x = 5

Divide both sides by 5: x = 1

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Write the verbal phrase below as an algebraic expression.

ra1l [238]

Answer:

9 x D

Step-by-step explanation:

D being the number of doughnuts produced.

8 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

Last 2 questions please help

Jlenok [28]

4.
Minimum: 20
First Quartile: 44
Median: 51
Third Quartile: 57
Maximum: 86

Ascending order : 20 41 44 44 49 53 55 57 80 86

6 0

2 years ago

The new athletic facility on the downtown campus measures 0.74 km by 4 km. How many square

masha68 [24]

Answer:

The sports facility is 2,960,000 sq. m in area.

Step-by-step explanation:

The dimension of the sport facility = 0.74 km by 4 km

The Length of the facility = 4 km

The width of the facility = 0.74 km

Now, 1 km = 1000 meters

⇒ 4 km = 4 x (1000) = 4000 meters

and 0.74 km = 0.74 x (1000) = 740 meters

Now, AREA OF THE RECTANGLE = LENGTH x WIDTH

So, the area of the facility = 4000 m x 740 km

= 2,960,000 sq. m

Hence, the sports facility is 2,960,000 sq. m in area.

7 0

3 years ago

Find the roots of parabola given by the following equation

riadik2000 [5.3K]

Answer:

3/2

-3

Step-by-step explanation:

here

5 0

3 years ago