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2 years ago

You have 40 minutes to exercise at the gym, and you want to burn 300 calories total using both

1 answer:

6 0

A and b are the minutes spent on machine A and machine B, respectively.

a+b = 40
b = 40-a

8a + 6b = 300
8a + 6(40-a) = 300
8a + 240 - 6a = 300
2a = 60
a = 30
b = 40-a = 10

30 minutes on machine A and 10 minutes on machine B

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PLS HELP ASAP ILL GIVE BRAINLKEST THANKS

stepladder [879]

Answer:

might b b let me re check

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

What is the value of (-14^0)^-2?

posledela

Answer:

The value of given expression is 1.

Step-by-step explanation:

The given expression is

$(-14^0)^{-2}$

If a is a real number, then $a^0=1$ .

$(-1)^{-2}$ $[\because a^0=1]$

Using exponent property we get,

$(-1)^{-2}=\frac{1}{(-1)^2}$ $[\because a^{-n}=\frac{1}{a^n}]$

$(-1)^{-2}=\frac{1}{1}$

$(-1)^{-2}=1$

Therefore the value of given expression is 1.

6 0

3 years ago

A boxplot helps to visualize the variability of a distribution. Five statistics form a boxplot, often referred to as the five-nu

ehidna [41]

Answer:

75%

Step-by-step explanation:

minimum data point to 1st quartile contains 25% of data

1st quartile to median contains 50% of data

median to 3rd quartile contains 75% of data

3rd quartile to maximum data point represents all of the data

6 0

2 years ago

Solve step by step 3(2x-4)=8

olganol [36]

Answer:

x = 3.33... or 10/3

Step-by-step explanation:

3(2x - 4) = 8

distribute the 3 into the parentheses

3 • 2x = 6x

3 • -4 = -12

6x - 12 = 8

add 12 to both sides

6x = 20

divide both sides by 6

x = 3.33... or 10/3

5 0

2 years ago

Read 2 more answers

“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