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4 years ago

Find the x- and y- intercepts of the line 6x-2y=24 write your answers as ordered pairs

Mathematics

1 answer:

MrRissso [65]4 years ago

7 0

Answer:

(4,0),(0,-12)

Step-by-step explanation:

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Kristin's grandparents started a savings account for her when she was born. They invested $500 in an account that pays 8% intere

Molodets [167]

Y = 500 + .08x is the correct answer

6 0

3 years ago

How much would $500 invested at 6% interest compounded monthly be worth after 5 years? round your answer to the nearest cent.

Lady bird [3.3K]

Use this formula: A = P(1 + r/n)^nt, where A is the amount after interest (what you are solving for), P is the amount you invested originally, r is the rate at which it was invested in decimal form, n is the number of times the compounding occurs each year, t is the time in years it is invested. It would look like this: A = 500(1 + [.06/12])^12*5. Do inside the parenthesis first to get 1 + .005 = 1.005. Now raise that to the 60th power (12 times 5 is 60) to get 1.34558. Now multiply that by the 500 out front to get a total amount of $674.43

8 0

3 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

4 years ago

5.4+ 3(52-15 +2)<br> What is it

iragen [17]

Answer:

122.4

Step-by-step explanation:

Remember PEMDAS

Parentheses

Exponents

Multiply

Divide

Add

Subtract

(52-15+2)=39

39x3=117

117+5.4=122.4

3 0

3 years ago

Read 2 more answers

What is the perpendicular adjacent sides in the triangle

d1i1m1o1n [39]

"Adjacent" means next to each other. "Perpendicular" means at an angle of 90 degrees.
<span>Rectangles (this includes squares) have adjacent perpendicular sides. So do right triangles.</span>

6 0

3 years ago

Read 2 more answers