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2 years ago

The oil company removed 5,421 liters of gas from a tank that contained 5,855 liters.

2 answers:

7 0

Liters of gas before they removed some gas = 5,855
Liters of gas AFTER they removed some gas = 5,421
To find how much liters of gas are left,
5,855 - 5,421 = the amount of liters left in the tank
The number of liters left in the tank is 434 Liters.
Oh oh! WAIT! Not done yet! The question asked for MELLILETERS, But we only have liters. <em>Let's convert it!

</em>1 Liter = 1000 ml
434 liters = 1000 × 434
= 434000 ml.

So, the answer to your question is <u>434000 ml</u>.
I hope this helps! ☺

dsp732 years ago

7 0

Answer:

434,000 mL. hope this helped!!

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Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side

lorasvet [3.4K]

Answer:

Step-by-step explanation:

cot x sec⁴ x = cot x+2 tan x +tan³x

L.H.S = cot x sec⁴x

=cot x (sec²x)²

=cot x (1+tan²x)² [ ∵ sec²x=1+tan²x]

= cot x(1+ 2 tan²x +tan⁴x)

=cot x+ 2 cot x tan²x+cot x tan⁴x

=cot x +2 tan x + tan³x [ ∵cot x tan x $=\frac{ \textrm{tan x }}{\textrm{tan x}}$ =1]

=R.H.S

(sin x)(tan x cos x - cot x cos x)=1-2 cos²x

L.H.S =(sin x)(tan x cos x - cot x cos x)

= sin x tan x cos x - sin x cot x cos x

$=\textrm{sin x cos x }\times\frac{\textrm{sin x}}{\textrm{cos x} } - \textrm{sinx}\times\frac{\textrm{cos x}}{\textrm{sin x}}\times \textrm{cos x}$

= sin²x -cos²x

=1-cos²x-cos²x

=1-2 cos²x

=R.H.S

1+ sec²x sin²x =sec²x

L.H.S =1+ sec²x sin²x

= $1+\frac{{sin^2x}}{cos^2x}$ [ $\textrm{sec x}=\frac{1}{\textrm{cos x}}$ ]

=1+tan²x $[\frac{\textrm{sin x}}{\textrm{cos x}} = \textrm{tan x}]$

=sec²x

=R.H.S

$\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}} = \textrm{2 csc x}$

L.H.S= $\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}}$

$=\frac{\textrm{sinx(1+cos x)+{\textrm{sinx(1-cos x)}}}}{\textrm{(1-cos x)\textrm{(1+cos x})}}$

$=\frac{\textrm{sinx+sin xcos x+{\textrm{sinx-sin xcos x}}}}{{(1-cos ^2x)}}$

$=\frac{\textrm{2sin x}}{sin^2 x}$

= 2 csc x

= R.H.S

-tan²x + sec²x=1

L.H.S=-tan²x + sec²x

= sec²x-tan²x

= $\frac{1}{cos^2x} -\frac{sin^2x}{cos^2x}$

$=\frac{1- sin^2x}{cos^2x}$

$=\frac{cos^2x}{cos^2x}$

8 0

3 years ago

A diagonals of a parallelogram measures 37 inches. The size measures 35 inches and 1 foot. Is the parallelogram a rectangle?

Lostsunrise [7]

No, a parallelogram is not a rectangle. meaning it has a completely different area formula, I know it sucks.

Formula: base × height

I don't use feet and inches. sorry I couldn't provide with to much help.

6 0

2 years ago

The regular price of an item is $150. The item is on sale at a discount rate of 30%.

Aleonysh [2.5K]

Answer:

$105

Step-by-step explanation:

30% of 150 is 45.

150-45 = 105

Hope that helps

5 0

2 years ago

Read 2 more answers

Estimate 13.23 - 2.570 by first rounding each number to the nearest tenth.

Temka [501]

Answer:

Step-by-step explanation:

13.23 = 10

2.570= 3

10-3=7

6 0

2 years ago

Read 2 more answers

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

2 years ago