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madam [21]
2 years ago
10

BRAINEST TO THE CORRECT ANSWER (50 points)

Mathematics
2 answers:
dmitriy555 [2]2 years ago
3 0

Answer:

(-4, -7)

Step-by-step explanation:

Equation 1:  5g - 2h = -6

Equation 2:  9 = h - 4g

Rewrite equation 2 to make h the subject:

9 = h - 4g

⇒  h = 9 + 4g

Substitute  h = 9 + 4g  into equation 1 and solve for g:

5g - 2(9 + 4g) = -6

⇒ 5g - 18 - 8g = -6

⇒ -3g = 12

⇒ g = -4

Substitute found value for g into either equation and solve for h:

9 = h - 4(-4)

⇒ 9 = h + 16

⇒ h = -7

Therefore, the correct solution is (-4, -7)

Tim has got the values for g and h mixed up.

Alex2 years ago
3 0

I guess question is incorrect itself.Assumed solution is provided

  • 59=2h-6k--(1)
  • 9k=h-49=>49=h-9k--(2)

Multiply 2 with eq(2)

  • 98=2h-18k--(3)

Subtract eq(3) from (1)

\\ \tt\hookrightarrow -39=12k

\\ \tt\hookrightarrow k=-39/12=-3.2

Put in eq(2)

\\ \tt\hookrightarrow 49=h+28.8\implies h=20.2

  • (h,k)=(20,2,-3.2)
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Write the standard equation of the circle with the given center is (0,0) and radius is 2.5
BlackZzzverrR [31]

Answer:

x² + y² = 6.25

Step-by-step explanation:

The standard equation of a circle is ( x – h)² + ( y - k)² = r² where the center is (h, k) and r is the radius.

From your given information that the center is (0, 0) then both h and k are 0. Then r is 2.5, so

( x – 0)² + ( y - 0)² = (2.5)²

x² + y² = (2.5)²

x² + y² = 6.25

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3 years ago
Dr black is standing 13 feet from a streetlamp. The lamp is making his shadow 9 feet long. He estimates that the angle of elevat
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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

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Butoxors [25]
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Shalnov [3]

Answer: 75.4 =

 70  

+ 5  

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Step-by-step explanation:

Hope this helps :)

8 0
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