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3 years ago

Please help! 3x^2 + 1/2 x 7y (When y = 3)

2 answers:

7 0

Factoring the equation:

3x^2 + 1/2 x 7(3)

3x^2 + 1/2 x 21

3 (x^2 + 1/2 x 21)

3 (x^2 + 7/2)

3 x 1/2 (2x^2 +7)

3/2 (2x^2 + 7)

mafiozo [28]3 years ago

4 0

Answer:

3x^2+21/2

Step-by-step explanation:

3x^2+1/2*7(3)

(a)=a so (3)=3

3x^2+1/2*7*3

1/2*7*3=21/2

3x^2+21/2

Hope this helps! Pls give brainliest!

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Answer: She took 6 yoga classes

Step-by-step explanation: So first

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3 years ago

A restaurant purchased kitchen equipment on January 1, 2017. On January 1, 2019, the value of the equipment was $14 comma 550

timama [110]

Answer:

$\frac{dV(t)}{dt} =$ - 1675.38

Step-by-step explanation:

In 2017, the vakue of the kitchen equipment was $14550

V(0)=$14550

Its value after then was modelled by $V(t)=14550e^{-0.158t$

We are required to find the rate of change in value on January 1, 2019

$V(t)=14550e^{-0.158t$

$\frac{dV(t)}{dt} =\frac{d}{dt}14550e^{-0.158t$

$\frac{dV(t)}{dt} =14550 \frac{d}{dt}e^{-0.158t$

$\\Let u= -0.158t,\frac{du}{dt}=-0.158$

$\frac{dV(t)}{dt} =14550 \frac{d}{du}e^u\frac{du}{dt}$

$\frac{dV(t)}{dt} =14550 X -0.158 e^{-0.158t}=-2298.9e^{-0.158t}$

In 2019, i.e. 2 years after, t=2

The rate of change of the value

$\frac{dV(t)}{dt} =-2298.9e^{-0.158X2}$

= $\frac{dV(t)}{dt} =-2298.9e^{-0.316}$ = - 1675.38

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3 years ago

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attashe74 [19]

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algol13

Answer:

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Answer:

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