1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
STatiana [176]
2 years ago
6

Mrs. Pham has 8 apples. She wants to give 3/4 of the

Engineering
2 answers:
Sergio039 [100]2 years ago
6 0
The answer would be 32/3 or if you want the simplified version it’s 10 2/3.
irakobra [83]2 years ago
6 0
6

2 = 1/4
4 = 2/4
6= 3/4
8=4/4
You might be interested in
You are hired as the investigators to identify the root cause and describe what should have occurred based on the following info
creativ13 [48]

Answer:

The mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.

Explanation:

Since density ρ = m/v where m = mass of fuel and v = volume of fuel, we need to find the mass of each volume of fuel.

So, m = ρv now ρ = specific gravity × density of water = 0.803 × 1000 kg/m³ = 803 kg/m³.

To find the mass of the 7,682 L of fuel, its volume is 7,682 dm³ = 7,682 dm³ × 1 m³/1000 dm³ = 7.682 m³.

It's mass, m = 803 kg/m³ × 7.682 m³ = 6168.646 kg

To find the mass of the extra 4,916 L of fuel added, we have

m' = ρv' where v' = 4,916 L = 4,916 dm³ = 4916 dm³ × 1 m³/1000 dm³ = 4.916 m³

m' =  803 kg/m³ × 4.916 m³ = 3947.548 kg

So, the total mass of the fuel is m" = m + m' = 6168.646 kg + 3947.548 kg = 10116.194 kg ≅ 10,166.2 kg

<u>Since this mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.</u>

4 0
2 years ago
one number is 11 more than another number. find the two number if three times the larger number exceeds four times the smaller n
vaieri [72.5K]

Answer:

a = 40

b = 29

Explanation:

Give a place holder for the numbers that we don't know.

Lets call the two numbers a and b.

From the given info, we can write an expression and solve it:

"one number is 11 more than another number"

a = 11 + b

from this, we know that a > b.

''three times the larger number exceeds four times the smaller number by 4"

3a = 4b + 4

Now we have 2 equations, we can use them to solve using whatever method you want.

a = 11 + b

3a = 4b + 4

I will be using matrices RREF to solve for this.

a - b = 11

3a - 4b = 4

\begin{bmatrix}1 & -1  & 11\\3 & -4 & 4 \end{bmatrix}

\begin{bmatrix}1 & 0  & 40\\0 & 1 & 29 \end{bmatrix}

a = 40

b = 29

6 0
3 years ago
By using order of magnitude analysis, the continuity and Navier-Stokes equations can be simplified to the Prandtl boundary-layer
Mademuasel [1]

Answer: Attached below is the well written question and solution

answer:

i) Attached below

ii) similar parameter =  \frac{V}{VoL } = 1 / Re

Explanation:

Using ;  L as characteristic length and Vo as reference velocity

i) Nondimensionalize the equations

ii) Identifying similarity parameters

the similar parameters are  = \frac{V}{VoL } = 1 / Re

Attached below is the detailed solution

7 0
2 years ago
A water reservoir contains 108 metric tons of water at an average elevation of 84 m. The maximum amount of electric energy that
zavuch27 [327]

Answer:

24.72 kwh

Explanation:

Electric energy=potential energy=mgz where m is mass, g is acceleration due to gravity and z is the elevation.

Substituting the given values while taking g as 9.81 and dividing by 3600 to convert to per hour we obtain

PE=(108*9.81*84)/3600=24.72 kWh

8 0
3 years ago
The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di
jeka57 [31]

Answer:

a) at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

b) daylight (d) = 0.50 μm

    Incandescent ( i ) =  1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

<em>Values are gotten from the table named: blackbody radiati</em>on functions

<u>a) Calculate the band emission fractions for the visible region</u>

at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

attached below is a detailed solution to the problem

<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>

For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0
2 years ago
Other questions:
  • What are some of the main causes of accidents?
    7·1 answer
  • According to information found in an old hydraulies book, the energy loss per unit weight of fluid flowing through a nozzle conn
    6·1 answer
  • The input power for a thermostat is wired to the
    8·1 answer
  • ITS FOR DRIVERS ED!!
    13·2 answers
  • How does refrigeration preserve food and dead bodies​
    12·2 answers
  • Select the correct answer.
    13·1 answer
  • Technician A says that in a worm gear steering system, most excessive steering free play is usually found in the gearbox. Techni
    13·1 answer
  • Some General Motors flex fuel vehicles do not use a fuel sensor to measure the percentage of ethanol in the fuel. These vehicles
    5·1 answer
  • Which is the correct way of dual dimensioning using the position method
    7·1 answer
  • What is the difference between class 1 and class 3 lever?
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!