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Dmitrij [34]
3 years ago
14

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

5 0
4 years ago
Solve The Inequality
sineoko [7]

Answer:

B

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
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

6 0
3 years ago
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Which is the graph of the linear inequality 2x – 3y < 12?
svp [43]

Answer:

see graph

Step-by-step explanation:

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2 years ago
In which situation do the quantities combine to make zero
Kaylis [27]

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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