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3 years ago

6

Simplify please help.

1 answer:

Vladimir [108]3 years ago

4 0

I hope this helps you

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The numbers 1 through 25 are written in red marker on slips of paper while the numbers 26 through 50 are written with blue marke

Gekata [30.6K]

A) $\frac{27}{50}$
b) <span> $\frac{3}{5}$ </span>

6 0

2 years ago

Y = 3x + 1<br> y = 3x - 3<br> Please help me out what’s x? <br> What’s y?

Harrizon [31]

Answer:

Step-by-step explanation:

use the comparison method

3x + 1 = 3x - 3

1 = -3

the expression is false for any value of x and y , so there no solution

4 0

3 years ago

The answer and steps for this question

melamori03 [73]

I will solve your system by substitution.<span><span>x=<span><span>3y</span>−1</span></span>;<span><span><span>5x</span>−<span>7y</span></span>=19</span></span>Step: Solve<span>x=<span><span>3y</span>−1</span></span>for x:Step: Substitute<span><span>3y</span>−1</span>forxin<span><span><span><span>5x</span>−<span>7y</span></span>=19</span>:</span><span><span><span>5x</span>−<span>7y</span></span>=19</span><span><span><span>5<span>(<span><span>3y</span>−1</span>)</span></span>−<span>7y</span></span>=19</span><span><span><span>8y</span>−5</span>=19</span>(Simplify both sides of the equation)<span><span><span><span>8y</span>−5</span>+5</span>=<span>19+5</span></span><span>(Add 5 to both sides)

</span><span><span>8y</span>=24</span><span><span><span>8y</span>8</span>=<span>248</span></span>(Divide both sides by 8)<span>y=3</span>Step: Substitute3foryin<span><span>x=<span><span>3y</span>−1</span></span>:</span><span>x=<span><span>3y</span>−1</span></span><span>x=<span><span><span>(3)</span><span>(3)</span></span>−1</span></span><span>x=8</span><span>(Simplify both sides of the equation)</span><span>

x=<span><span>8<span> and </span></span>y</span></span>=3

6 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

Owen had 3 different kinds of stickers that he wanted to put on paper he put a bird sticker on every 30th paper a sports sticker

sveta [45]

Do 60+50+30 to get the answer

3 0

2 years ago

Read 2 more answers