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3 years ago

Can someone help me please?

Mathematics

1 answer:

Bond [772]3 years ago

6 0

Answer:

25, 29

Step-by-step explanation:

The Pythagorean Theorem is a² + b² = c² where a and b are legs and c is the hypotenuse. In this case, a = 2, b = 5 and we are solving for c, which is also known as AC here. Therefore,

4 + 25 = AC²

AC² = 29

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Write the first five terms of the sequence in which the nth term is an=(n+1)! / n+1

goblinko [34]

Answer:

1, 2, 6, 24, 120

Step-by-step explanation:

let's work on each of the five terms requested for the sequence. Notice that the first term is when we give n the value 1 $a_1=(1+1)!/(1+1)=2!/2=2/2=1$

Next term (when n=2) is: $a_1=(2+1)!/(2+1)=3!/3=3*2/3=2$

Next term (when n=3) is: $a_1=(3+1)!/(3+1)=4!/4=4*3*2/4=6$

Next term (when n=4) is: $a_1=(4+1)!/(4+1)=5!/5=5*4*3*2/5=24$

Next term (when n=5) is: $a_1=(5+1)!/(5+1)=6!/6=6*5*4*3*2/6=120$

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4 years ago

Solve The Inequality

sineoko [7]

Answer:

Step-by-step explanation:

I just solved it now

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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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Answer:

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Step-by-step explanation:

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Answer:

an integer plus its opposite sum zero. The opposite of a number is called the additive inverse because the two numbers' sum is zero.

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