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2 years ago

What is the change in elevation from 8500m and -300m

Mathematics

1 answer:

bogdanovich [222]2 years ago

5 0

Answer:

8800m

Step-by-step explanation:

difference between 8500 and -300 is 8800m

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3 years ago

I have a geometry test I’m gonna fail. If you know any of these answers please put the number of the question,than the answer. I

blsea [12.9K]

Answer:

Step-by-step explanation:

20) Angles 1 and 2 are supplementary angles( sum of angle 1 and angle 2 is 180 degrees)

21) Angles 1 and 2 are supplementary angles( sum of angle 1 and angle 2 is 180 degrees)

22) Angle TUW + Angle WUV =Angle TUV

Therefore

(7x-9) + (5x - 11) = (9x+1)

7x-9 + 5x - 11 = 9x+1

7x +5x-9x =1 + 11 + 9

3x = 21

x = 21/3= 7

23) GEF - 13 = 5DEG

and DEF = 149 degrees

GEF = 5DEG + 13

Sum of angles (GEF + DEG)=DEF

Therefore,

DEG + 5DEG - 13 = 149

6DEG = 149 + 13 = 162

DEG = 162/6 = 27

GEF = 5DEG = 27 × 5 =135

24) 7x - 1 + 6x -1 = 180(sum of angles in a straight line)

13x-2 = 180

13x = 180+2 = 182

x = 182/13 = 14

25) 5x + 4 + 8x - 7 = 180( sum of supplementary angles is 180)

13x = 180+7-4= 183

x = 183/13 = 14.08

3 0

2 years ago

“encontrar la integral indefinida y verificar el resultado mediante derivación”

Oliga [24]

$I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx$

Haz la sustitución:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C$

Para confirmar el resultado:

$\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}$

$I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx$

Sustituye:

$y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx$

$\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C$

(Te dejaré confirmar por ti mismo.)

$I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx$

Sustituye:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C$

$I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}$

Sustituye:

$u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}$

$\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C$

Podemos hacer que esto se vea un poco mejor:

$\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}$

$\implies I=-\dfrac{(t+1)^4}{4t^4}+C$

4 0

3 years ago

4y-4=_____<br> 4(y+1)=___<br> 2+4y=____

34kurt

If they are looking to have them all simplified, this is how they should look.

4y - 4
4y + 4
4y + 1

3 0

3 years ago