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3 years ago

The quotient of a number and four?

Mathematics

1 answer:

BARSIC [14]3 years ago

5 0

Answer:

$\frac{n}{4}$

Step-by-step explanation:

Since the quotient is the result of the division of two terms, we should end up with a fraction, which represents division.

To start, we'll have to pick a letter to represent the number referenced in the question simply as "a number". I picked $n$ .

The quotient of $n$ and four would be $n$ divided by four. This <em>could</em> be written as $n$ ÷ 4, but I think it works better as a fraction:

$\frac{n}{4}$

Therefore, the quotient of a number and four, where the number is represented by $n$ is $\frac{n}{4}$ .

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50 parents were asked which activities they enrolled their infant in.

Salsk061 [2.6K]

Answer: gymnastics 35

music 21

swimming 36

Step-by-step explanation:

gymnastics 35=20+9+6

music 21=8+7+6

swimming 36=20+9+7

4 0

3 years ago

What is the y-value of the solution to the system of equations?

motikmotik

Answer:

{x = -3 , y=2 (Isolved for both variables be elimination)

Step-by-step explanation:

Solve the following system:

{3 x + 5 y = 1 | (equation 1)

7 x + 4 y = -13 | (equation 2)

Swap equation 1 with equation 2:

{7 x + 4 y = -13 | (equation 1)

3 x + 5 y = 1 | (equation 2)

Subtract 3/7 × (equation 1) from equation 2:

{7 x + 4 y = -13 | (equation 1)

0 x+(23 y)/7 = 46/7 | (equation 2)

Multiply equation 2 by 7/23:

{7 x + 4 y = -13 | (equation 1)

0 x+y = 2 | (equation 2)

Subtract 4 × (equation 2) from equation 1:

{7 x+0 y = -21 | (equation 1)

0 x+y = 2 | (equation 2)

Divide equation 1 by 7:

{x+0 y = -3 | (equation 1)

0 x+y = 2 | (equation 2)

Collect results:

Answer: {x = -3 , y=2

5 0

3 years ago

Read 2 more answers

Answer asap please will mark as brainlist

Leona [35]

Answer:

Exact Form:

√ 30

Decimal Form:

5.47722557 …

Step-by-step explanation: i would go with D

6 0

3 years ago

Which ordered pair is a solution to this equation?

ElenaW [278]

For (11, 1): (11 + 3)1 = 14(1) = 14
For (5, 2): (5 + 3)2 = 8(2) = 16 ≠ 14
For (7, 2): (7 + 3)2 = 10(2) = 20 ≠ 14
For (3, 2): (3 + 3)2 = 6(2) = 12 ≠ 14

Therefore, (11, 1) is a solution to this equation.

8 0

3 years ago

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago

The quotient of a number and four?​

The quotient of a number and four?