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3 years ago

Find the perimeter and area of each figure below

Mathematics

1 answer:

Olin [163]3 years ago

6 0

For b it's 39 because you multiply 5 and 7 giving you 35 then add 4 Wich equals 39. I might be incorrect. check it out.

You might be interested in

Name five fractions whose values are between 1/2and<br> 7/8

pantera1 [17]

Answer:

9/16 10/16 11/16 12/16 13/16

Step-by-step explanation:

7 0

3 years ago

What are the coordinates of Point A after the

Flura [38]

Answer:

A' (-3,7)

B' (0,10)

C' (12,10)

D' (-2,-4)

Step-by-step explanation:

A' (-6+3 , 3+4) = (-3,7)

B' (-3+3 , 6+4) = (0,10)

C' (9+3 , 6+4) = (12,10)

D' (-5+3 , -8+4) = (-2,-4)

5 0

3 years ago

A quadratic functions has zeros of 5 and -2. What could be the equation in factored form?

marusya05 [52]

Hi,

A is the answer as : -2+2 = 0 and 5 -5 = 0

8 0

3 years ago

How would you use estimation to evaluate this expression 10.2 x [(2x3.7)+8]

Goryan [66]

Round off the decimals to a sensible number. So 3.7 would be 4 and 10.2 would be 10. In this case, it would be (2 × 4 + 8) × 10 which is 160.

6 0

3 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