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2 years ago

3

The graph of f(x) = 0.6x is replaced by the graph of g(x) = 0.6x + k. If g(x) is obtained by shifting f(x) up by 6 units, then w

hat is the value of k?
A.=-6
B.=-1/6
C.=1/6
D.=6

2 answers:

elena55 [62]2 years ago

4 0

Answer:

k = 6

Step-by-step explanation:

i got it right on the test and i also graphed it

Please mark brainliest!

by the way when you need to graph something use a website called Desmos. or just go onto the search bar on the top of your screen adn type in desmos. com

AleksAgata [21]2 years ago

3 0

Answer:

D

Step-by-step explanation:

Given the graph of f(x), then

f(x) + k represents a vertical translation ( shift ) of k units

• If k > 0 then shift of k units up ↑

• If k < 0 then shift of k units down ↓

Here g(x) is obtained by shifting f(x) up by 6 units

here k = 6 > 0

hence g(x) = 0.6x + 6 → D

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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

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

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

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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)

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

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

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

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$\Rightarrow A= 92 \times 2$

⇒A= 184.

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