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3 years ago

Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.

Mathematics

2 answers:

pickupchik [31]3 years ago

8 0

Answer:

the answer to the question is 13x

valentina_108 [34]3 years ago

4 0

Answer : The sum of these quotients is, 13x

Step-by-step explanation :

As we are given that:

Quotients are, $\frac{25x^2}{5x}$ and $\frac{8x^2}{x}$

Now we have to determine the sum of these quotients.

Sum of quotients = $\frac{25x^2}{5x}+\frac{8x^2}{x}$

Now we are taking LCM.

Sum of quotients = $\frac{25x^2+40x^2}{5x}$

Sum of quotients = $\frac{65x^2}{5x}$

Sum of quotients = 13x

Therefore, the sum of these quotients is, 13x

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-8x+32=4(-2-4x)<br>What does X equal

marshall27 [118]

Answer:

x = - 5

Step-by-step explanation:

Given

- 8x + 32 = 4(- 2 - 4x) ← distribute parenthesis on right side

- 8x + 32 = - 8 - 16x ( add 16x to both sides )

8x + 32 = - 8 ( subtract 32 from both sides )

8x = - 40 ( divide both sides by 8 )

x = - 5

7 0

3 years ago

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ladessa [460]

24 is the answer welcome !

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3 years ago

1) Find the Range of the following number set: (22,7, 22, 1, 7, 18, 18, 16, 6, 6} 2 points

lisabon 2012 [21]

Answer:

RANGE=21

Step-by-step explanation:

Least to greatest is KEY!

7 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

Work out the value of each expression, when r= 2.5 and t = 6.

Usimov [2.4K]

Answer:

a) 4(t - r) = 4(6 - 2.5) = 4 x 3.5 = 14

b) 4t - r = (4 x 6) - 2.5 = 24 - 2.5 = 21.5

c) 2(3t - 10) = 2(3 x 6 - 10) = 2(18 - 10) = 2 x 8 = 16

d) (t - 2)² = (6 - 2)² = 4² = 4 x 4 = 16

e) cannot decipher the equation

f) 6r + t = 6 x 2.5 + 6 = 15 + 6 = 21

8 0

2 years ago

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