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3 years ago

PLEASE HELP THIS IS MY LAST QUESTIONNNN

Mathematics

2 answers:

Vinil7 [7]3 years ago

7 0

Answer:

500 kilowatt-hours.

Step-by-step explanation:

Let k = number of kilowatt-hours.

$0.15k+30=105\\0.15k=75\\k=500$

Therefore, Dalton used 500 kilowatt-hours of electricity.

Ray Of Light [21]3 years ago

6 0

5 he used 5 hours or however it is out but it’s “5”

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Find the volume of a trough 5 meters long whose ends are equilateral triangles, each of whose

yawa3891 [41]

Answer:

$V=5\sqrt{3}\ m^3$

Step-by-step explanation:

we know that

The volume of a trough is equal to

$V=BL$

where

B is the area of equilateral triangle

L is the length of a trough

step 1

Find the area of equilateral triangle B

The area of a equilateral triangle applying the law of sines is equal to

$B=\frac{1}{2} b^{2} sin(60\°)$

where

$b=2\ m$

$sin(60\°)=\frac{\sqrt{3}}{2}$

substitute

$B=\frac{1}{2}(2)^{2} (\frac{\sqrt{3}}{2})$

$B=\sqrt{3}\ m^{2}$

step 2

Find the volume of a trough

$V=BL$

we have

$B=\sqrt{3}\ m^{2}$

$L=5\ m$

substitute

$V=(\sqrt{3})(5)$

$V=5\sqrt{3}\ m^3$

3 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

2 years ago

What is SOH CAH TOA?

vitfil [10]

Answer:

Soh= sin= opp/hyp cah= cos=adj/hyp

toa= tan= opp/adj

Step-by-step explanation:

4 0

2 years ago

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Which solid could have exactly two planes of symmetry?

kodGreya [7K]

Answer:

Triangular Prism

Step-by-step explanation:

A plane of symmetry divides a three dimensional shape into two congruent halves that are mirror images of each other. An isosceles triangular based prism has 2 planes of symmetry. An isosceles triangular based prism has 2 planes of symmetry. An isosceles triangular based prism has 2 planes of symmetry.

4 0

1 year ago

Angles α and β are the two acute angles in a right triangle, where α = 5x 3 + 20 and β = 2x 3 + 14. Find α.

Tom [10]

The acute angles of a right triangle are complementary, so
α + β = 90
(5x/3 +20) + (2x/3) + 14) = 90 . . . . . . substitute given values
7x/3 +34 = 90 . . . . . . . . . . . . . . . . . . . collect terms
7x/3 = 56 . . . . . . . . . . . . . . . . . . . . . . . subtract 34
x = (3/7)*56 = 24 . . . . . . . . . . . . . . . . . multiply by 3/7

Then the value of α is ...
α = 5*24/3 +20 = 60

7 0

3 years ago

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