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3 years ago

8

Brainliest if correct

2 answers:

Bad White [126]3 years ago

8 0

Answer:

(-2,11) (0,3) (2,-5) (4,-13)

Step-by-step explanation:

used desmos calculator

Nookie1986 [14]3 years ago

3 0

Answer: (-2,11) (0,3) (2,-5) (4,-13)

<em>i got it now i thought i was gone be duumd to not do it LOLL Buh here you got brainliest ?>33 ~shay</em>

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Roger has $5000 more invested at 7 1/4% then he has at 6%. If his total profit at the end of the year $659.30, how much has he i

kkurt [141]

Answer:

You have the number of nickels and the number of quarters both expressed in terms of the number of dimes. So let your variable be the number of dimes. Call this D. Then the number of nickels is 4D, and the number of quarters is D+6. The total value is

0.10(D) + 0.05(4D) + 0.25(D+6) = 4.25

Solve this for D, and use D to find 4D and D+6.

Step-by-step explanation:

4 0

3 years ago

Someone help me ASAP

MA_775_DIABLO [31]

Okay so if you add the top and the

3 0

3 years ago

Find the point on the terminal side of θ = 3pi/4 that has a y coordinate of 1.

natima [27]

Tanθ = tan(3pi/4) = -1
y/x = -1
1/x = -1
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Point is (-1, 1).

4 0

3 years ago

The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

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2 years ago

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Factor the four-term polynomial.<br> pq - 2r + pr-29

vredina [299]

Your answer should be -18?

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3 years ago