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katrin2010 [14]
3 years ago
8

Which system of equations can be used to find the roots of the equation 4 x Superscript 5 Baseline minus 12 x Superscript 4 Base

line + 6 x = 5 x cubed minus 2 x?
StartLayout Enlarged left-brace 1st Row y = negative 4 x Superscript 5 Baseline + 12 x Superscript 4 Baseline minus 6 x 2nd row y = 5 x cubed minus 2 x EndLayout
StartLayout Enlarged left-brace 1st Row y = 4 x Superscript 5 Baseline minus 12 x Superscript 4 Baseline + 5 x cubed + 4 x 2nd row y = 0 EndLayout
StartLayout Enlarged left-brace 1st row y = 4 x Superscript 5 Baseline minus 12 x Superscript 4 Baseline + 6x 2nd row y = negative 5 x cubed + 2 x EndLayout
StartLayout Enlarged left-brace 1st row y = 4 x Superscript 5 Baseline minus 12 x Superscript 4 Baseline + 6x 2nd row y = 5 x cubed minus 2 x EndLayout
Mathematics
1 answer:
ehidna [41]3 years ago
3 0

Answer:

D. on Edg 2020

Step-by-step explanation:

You are welcome, I helped better than that other guy!!!

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Read 2 more answers
Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
Viktor [21]

Answer:

Therefore k= \frac{ln2 }{18}, A=184

Step-by-step explanation:

Given function is

T(t)=230 -e^{-kt}

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

138=230 -Ae^{-k\times 18}

\Rightarrow  -Ae^{-18k}=138-230

\Rightarrow  Ae^{-18k}=92 .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

184=230-Ae^{-k\times 36}

\Rightarrow Ae^{-36k}=230-184

\Rightarrow Ae^{-36k}=46.......(2)

Dividing (2) by (1)

\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}

\Rightarrow e^{-18k}=\frac{46}{92}

Taking ln both sides

ln e^{-18k}=ln\frac{46}{92}

\Rightarrow -18k =ln (\frac12)

\Rightarrow -18k= ln1-ln2

\Rightarrow k= \frac{ln2 }{18}

Putting the value k in equation (1)

Ae^{-18\frac{ln2}{18}}=92

\Rightarrow A e^{ln2^{-1}}=92

\Rightarrow A.2^{-1}=92

\Rightarrow \frac{A}{2}=92

\Rightarrow A= 92 \times 2

⇒A= 184.

Therefore k= \frac{ln2 }{18}, A=184

7 0
3 years ago
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