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2 years ago

15

Please help with these linear equations

1 answer:

Ann [662]2 years ago

5 0

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onathan bought a new computer for $2,016, using the electronics store's finance plan. He will pay $112 a month for 18 months. Wh

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The money he still owes=$2016-122x

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3 years ago

Use the Chain Rule to find dw/dt.<br> w = xey/z, x = t3, y = 7 − t, z = 8 + 4t<br> dw<br> dt<br> =

Hitman42 [59]

$\frac{dw}{dt} = \frac{dx}{dt} \times \frac{dw}{dx} + \frac{dy}{dt} \times \frac{dw}{dy} + \frac{dz}{dt} \times \frac{dw}{dz} \\ \frac{dw}{dt} = 3t ^{2} \times {e}^{ \frac{y}{z} } + - t \times \frac{x {e}^{ \frac{y}{z} } }{z} + 4 \times - \frac{xy {e}^{ \frac{y}{z} } }{ {z}^{2} } \\ \frac{dw}{dt} = \frac{ {e}^{ \frac{y}{z} } (3 {t}^{2} {z}^{2} - tzx - 4xy)}{ {z}^{2} }$

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2 years ago

Find the amount in an account if 7,650 is invested at 9.15 percent compounded quarterly for 8 years and 6 months

storchak [24]

We have been given that in an account an amount of 7,650 is invested at 9.15 percent compounded quarterly for 8 years and 6 months.

We will use compound interest formula to find our answer.

$A=Pcdot(1+\frac{r}{n}) ^{nT}$ ,

Where, P= principle amount, A= amount after T years, n= period of compounding and r = interest rate (decimal).

Let us substitute our given values in our formula.

$A=7650(1+\frac{0.0915}{4} )^{(8.5\times4)}$

$A=7650(1+0.022875)^{34}$

$A=7650(1.022875)^{34}$

$A=7650\cdot 2.15758136398$

$A=16505.4974$

Therefore, after 8 years and 6 months our amount will be 16505.497.

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2 years ago

If 3b+2-6b=32+5b-2b, find b

natulia [17]

I believe the answer is b= -5

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2 years ago

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4vir4ik [10]

Answer:

I think It would be A but I'm not exactly sure

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3 years ago

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