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
UkoKoshka [18]
3 years ago
7

How many of each type of box is it carrying?

Mathematics
2 answers:
CaHeK987 [17]3 years ago
8 0

Answer:

Number of large boxes: 55

Number of small boxes:65

Step-by-step explanation:

Let be "l" the number of large boxes and "s" the number of small boxes.

Set up a system of equations:

\left \{ {{l+s=120} \atop {60l+30s=5250}} \right.

Use the methof of elimination. Mulitply the first equation by -60 and add both equations. Then solve for "s":

\left \{ {{-60l-60s=-7200} \atop {60l+30s=5250}} \right.\\.........................\\-30s=-1950\\s=65

Substitute s=65 into any of the original equations and solve for "l":

l+65=120\\l=55

garri49 [273]3 years ago
6 0

<u>Answer:</u>

Number of large boxes = 55

Number of small boxes = 65

<u>Step-by-step explanation:</u>

We know that large boxes weight 60 pounds each while small boxes weight 30 pounds each and they are 120 boxes in total which weigh 5250 pounds in total.

Assuming l to the number of large boxes and s to be the number of small boxes, we can write the following equations:

l+s=120 --- (1)

60l+30s=5250 --- (2)

From equation (1):

l=120-s

Substituting this value of l in (2):

60(120-s)+30s=5250

7200-60s+30s=5250

60s-30s=7200-5250

30s=1950

s = 65

Now finding the value of s:

l=120-65

l = 55

Therefore, number of large boxes = 55 and number of small boxes = 65.

You might be interested in
Terry Bergolt's bank granted him a single-payment loan of $4,400 at an interest rate of 6% exact interest. The term of the loan
Mashutka [201]
The correct answer is (b)
6 0
3 years ago
Read 2 more answers
Which of the relations has a domain of {-5, 0, 5}?
STALIN [3.7K]
B is the correct answer i believe
8 0
3 years ago
Read 2 more answers
Which shows the expression below in simplified form? (3.7 × 1013) + (4.5 × 1012) 4.15 × 1013 8.2 × 1013 3.745 × 1012 8.2 × 1012
frosja888 [35]
What is your first expression? because i can not answer without the first expression.
8 0
3 years ago
A survey reveals 252 out of 350 students in 7th grade like to read. What percent like to read? Write a percent proportion equati
ddd [48]

Answer

72% students likes to read .

Step-by-step explanation:

Let us assume that the percent of student likes to read be x.

As given

A survey reveals 252 out of 350 students in 7th grade like to read.

As Percent is always from 100 percent .

Thus

x is from 100 in case of proportions.

252 students are from 350 students .

Therefore the proportion equations becomes

x : 100 :: 252 : 350

\frac{x}{100}= \frac{252}{350}

Simplify the above

x= \frac{252\times 100}{350}

x= \frac{25200}{350}

x = 72%

Therefore 72% students likes to read .

6 0
3 years ago
An object moving along a curve is at position P(2cost, sint) where t denotes time in seconds. Find the speed of the object at po
shepuryov [24]
We have 2 equations to specify the location of the object and we desire the velocity. In order to get that, we simply need to calculate the first derivative of each location equation. So:    X = 2 cos(t)  X' = 2 (-sin(t))  X' = -2 sin(t)    Y = sin(t)  Y' = cos(t)    So the velocity vector at time t is (-2sin(t), cos(t)). But you want the velocity. So using the Pythagorean theorem we can get that by calculating the square root of the sum of the squares. So:  V = sqrt((-2sin(t))^2 + cos^2(t))  V = sqrt(4sin^2(t) + cos^2(t))    Speed at t = 1, is  V = sqrt(4sin^2(1) + cos^2(1))  V = sqrt(2.832293673 + 0.291926582)  V = sqrt(3.124220255)  V = 1.767546394    And t=3:  V = sqrt(4sin^2(3) + cos^2(3))  V = sqrt(0.079659427 + 0.980085143)  V = sqrt(1.05974457)  V = 1.029438959    Now asking for velocity as a function of P, we have a bit of a complication. As shown above, it's trivial to calculate velocity as a function of t. But if all you're given is the X and Y coordinates of the object, we have a bit more work to do. The below equations will be using the trigonometric identity of cos^2(a) + sin^2(a) = 1 for any angle a.    X = 2 cos(t)  X' = -2 sin(t)  We want to get from X which is 2cos(t) to X'^2 which is 4sin^2(t). So:  X/2; We now have cos(t)  (X/2)^2: We now have cos^2(t)  1-(X/2)^2: We now have sin^2(t)  4(1-(X/2)^2): We now have 4sin^2(t) which is what we want. Time to simplify  4(1 - (X/2)^2)  4(1 - (X^2/4))  4 - 4(X^2/4)  4 - X^2    Now we need to get from Y to Y'^2. Will do the same as for X to X'^2, but without all the comments.  Y = sin(t)  Y' = cos(t)  Y'^2 = 1 - Y^2    So the equation for the velocity as a function of X,Y we get  V = sqrt(4 - X^2 + 1 - Y^2)  V = sqrt(5 - X^2 - Y^2)    In summary:  Position at time t = (2cos(t), sin(t))  Velocity vector at time t = (-2 sin(t), cos(t))  Velocity as function of t is: V = sqrt(4sin^2(t) + cos^2(t))  Velocity as function of P is: V = sqrt(5 - X^2 - Y^2)  Is object traveling at constant speed? NO  Velocity at t = 1 is: V = 1.767546394  Velocity at t = 2 is: V = 1.029438959
7 0
4 years ago
Other questions:
  • <img src="https://tex.z-dn.net/?f=%20-%20%20%5Cfrac%7B5%7D%7B8%7D%20%20%2B%20%20%5Cfrac%7B1%7D%7B4%7D%20" id="TexFormula1" title
    10·1 answer
  • Solve-<br> 10x + 9 = 4x - 9
    13·2 answers
  • Lesson 6.2: applications of linear systems this is holding me back from completing the course. Please help me guys and girls.
    8·2 answers
  • What is 1/3 of 1/2?
    7·2 answers
  • PLS HELP ASAP WILL GIVE BRAINlEST AND 100 POINTS
    11·1 answer
  • PLEASE HELP MY FRIEND SAYSNTHIS IS HIGHSCHOOL MATH AND IM IN 7TH AHH HELP ASAP
    10·1 answer
  • What expression is equal to (6i - 5) - (2 - 3i)
    9·2 answers
  • Use 10% to find 90% of 170
    5·2 answers
  • Can you please help me with this question
    9·1 answer
  • Can anyone help me i really need it rn<br>​
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!