All categories

3 years ago

Solve this equation, -13m=-377

Mathematics

1 answer:

lozanna [386]3 years ago

5 0

-13m = -377
m= -377/-13
m=29
the negatives cancel out. so just do 377/13 which equals 29

You might be interested in

For what values of θ, −π ≤ θ ≤ π, is y = tan θ undefined?

pogonyaev

Answer:

The answer is II and IV ⇒ the 3rd answer

Step-by-step explanation:

* lets check the domain of angle Ф

∵ -π ≤ Ф ≤ π

∵ y = tanФ

∵ tanФ = sinФ/cosФ

* The effective term which makes tanФ undefined is cosФ

- If cosФ = 0, than tanФ will be undefined

* Lets check the angles that have cosФ = 0

∵ The unit circle intersect x-axis at point (1 , 0) and (-1 , 0)

∵ The unit circle intersect y-axis at point (0 , 1) and (0 , -1)

∵ cosФ = x-coordinates of the points

∵ The points of intersection with the y-axis have x- coordinates = 0

∴ The angles on the y-axis have cosФ = 0

* The angles on the +ve part of y-axis are π/2 and -3π/2

The angles on the -ve part of y-axis are -π/2 and 3π/2

∴ The tan of π/2 , 3π/2 , -π/2 , -3π/2 undefined

* In the problem

I. -π ⇒ defined ⇒ on the -ve part of x-axis

II. -π/2 ⇒undefined

III. 0 ⇒ defined ⇒ on the +ve part of the x-axis

IV. π/2 ⇒ undefined

V. π ⇒ defined ⇒ on the -ve part of x-axis

∴ The answer is II and IV ⇒ the 3rd answer

3 0

3 years ago

I WILL GIVE YOU BRAINLIEST IF YOU ANSWER CORRECTLY.........Please help

sammy [17]

$0.0008 = 8 \times 10 {}^{ - 4} \\ \\ \frac{8 \times 10 {}^{ - 4} }{2 \times 10 {}^{ - 6} } \\ \\ 4 \times 10 {}^{6 - 4} \\ \\ 4\times 10^2$

4 0

3 years ago

A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}$

(i) Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}$

(ii) Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C$

(iii) Find the constant of integration

$\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)$

(iv) Solve for A as a function of time.

$\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}$

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

$\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}$

b.Calculate the concentration

$A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}$

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is

$\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}$

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

$\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}$

(c) Concentration at infinite time

$\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}$

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

3 0

3 years ago

In July 15 days of rain, 10 days without rain, 6 days of only sunshine. What fraction of the days in July did not have any rain?

KonstantinChe [14]

Answer:

16/31 days

Step-by-step explanation:

Add up all the days: 15+10+6=31

Add up the days with no rain: 10+6=16

With no rain as numerator and total as denominator you get an answer of 16/31

5 0

3 years ago

Read 2 more answers

Janet drove 420 miles in 6 hours. What was her constant rate of change?<br><br><br> Pls answer

Ira Lisetskai [31]

Answer:

70 miles per hour

Step-by-step explanation:

5 0

2 years ago