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Vlad1618 [11]
3 years ago
15

What is the answer to 6(x+1)-5x=8+2(x-1)

Mathematics
1 answer:
Klio2033 [76]3 years ago
8 0

Answer:

answer is X=0

<em><u>mark this as brainiest</u></em>.

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Drag each tile to the correct box.
Tatiana [17]
V= 3.14 r2 h
1. 3.14 x 14 x 14 x 18 = 11083.54
2. 3.14 x 13 x 13 x 17 = 9025.8
3. 3.14 x 15 x 15 x 14 = 9896.02
4 .3.14 x 12 x 12 x 20 = 9047.79
so the order would be the second then the forth then the third then the first
4 0
4 years ago
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la siguinete tabla mustran la correspondencia entre dos variables. Determina si es funcion o relacion
Vaselesa [24]

Answer:

where's the table. on és la taula?

Step-by-step explanation:

6 0
3 years ago
If 3x^2 + y^2 = 7 then evaluate d^2y/dx^2 when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -,
S_A_V [24]
Taking y=y(x) and differentiating both sides with respect to x yields

\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0

Solving for the first derivative, we have

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y

Differentiating again gives

\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0

Solving for the second derivative, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}

Now, when x=1 and y=2, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63
3 0
3 years ago
Find the area of the sector (shaded part). Round to the nearest hundredth! <br><br> No links!!
igor_vitrenko [27]

Answer:

~183.78

Step-by-step explanation:

A = pi 9² ≈254.47

260/360=13/18

254.47/18×13≈183.78

6 0
3 years ago
Integral Calculation
Tatiana [17]

Answer:

\frac{4}{3(e^2+1)}

Step-by-step explanation:

We want to evaluate:

\int\limits^1_{-1} {\frac{x^2+1}{e^2+1} } \, dx

This is the same as:

\frac{2}{e^2+1} \int\limits^1_{0} {x^2+1} \, dx

We integrate to obtain:

\frac{2}{e^2+1} (\frac{x^3}{3}+x)|_0^1

We evaluate to obtain:

\frac{2}{e^2+1} (\frac{1^3}{3}+1)=\frac{4}{3(e^2+1)}

8 0
3 years ago
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