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2 years ago

Help please with algebra homework

Mathematics

1 answer:

Hunter-Best [27]2 years ago

7 0

1)

1: It's a function
2: It's not a function
3: It's a function

4)

4: It's not a function
5: It's a function
6: It's a function

Hope I did it right and helped you ^.^

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Choose the name of the linear equations form given:

pickupchik [31]

None of the above. The first two are slope-intercept, whereas the third one isn't a form.

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3 years ago

One of the x-intercepts of the parabola represented by the equation y = 3x^2 + 6x − 10 is approximately (1.08, 0).

Igoryamba

3x^2 + 6x - 10 = 0

x = [-6 +/- sqrt(^2 - 3*3*-10)] / 2*3

= 1.08 and -3.08

Other x -intercept is (-3.08,0)

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3 years ago

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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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In the rhombus m 1=15x, m 2=x+y, m 3=30z Find The Value of each variable

Sonbull [250]

First one m = .0666666
second one i don't know
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3 years ago

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Graph the numbers that are solutions to both inequalities below.

aleksklad [387]

Answer:

3x>12 2x/3 ≤ 6

3x>12, x = 4 2x/3, x = 2 ≤ 6

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2 years ago