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
umka2103 [35]
3 years ago
15

16. (a + b)(3a - b)(2a +76)

Mathematics
1 answer:
balu736 [363]3 years ago
7 0
6a^3+228a^2+4a^2b+152ab-2ab-76b^2
You might be interested in
Find the midpoint of the line segment connecting (1,1) and<br> (3,-3).
likoan [24]

Answer:

2,-1

Step-by-step explanation:

graph, find length between points (sqrt 20) and slope (-2). sqrt20/2=sqrt5.

change in y/change in x = -2, change in y squared + change in x squared= 5. solve system of equations to get x=1, y=-2, or x=-1, y=2. i think i did something wrong but the answer should be correct

8 0
3 years ago
At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel
Alexeev081 [22]
Use Law of Cooling:
T(t) = (T_0 - T_A)e^{-kt} +T_A
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
42 = (80-20)e^{-3k} +20 \\  \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\  \\ k = -\frac{1}{3} ln (\frac{11}{30})

Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3}  = 60 \\  \\  (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\  \\  \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\  \\ \\  t = 4.959

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
8 0
3 years ago
Hello plz solve this question whoever solves best gets brainlest and points any links will be reported thank you
Drupady [299]

Answer:

D is correct

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The area of a room in a dollhouse is 1,248 square inches. The width of the room is 8 inches. How long is the room?
AveGali [126]
The length of the dollhouse is 156 square inches
6 0
3 years ago
The inverse of the function graphed below is a function.?<br> A. True<br> B. False
Zinaida [17]

Step-by-step explanation:

A,true

doesn't interfere with any function restriction

3 0
3 years ago
Other questions:
  • Expressions equivalent to 6x+1-(3x-1)
    9·1 answer
  • A group of freshmen at a local university consider joining the equestrian team. Thirty‑five percent of students choose Western r
    7·1 answer
  • The Super Shop is having a grand opening sale. Today's special includes 4
    11·1 answer
  • 6. 150 is 75% of what number?
    7·2 answers
  • Determine whether the conclusion is based on inductive or deductive reasoning.
    8·1 answer
  • A man age is three times his sons age. ten years ago was five times his sons age. find their current age.
    15·1 answer
  • Is it possible to have a bisector that is not perpendicular to the segment that it bisects?
    15·1 answer
  • 1. Write an integer for a loss of 14 points.
    8·2 answers
  • A teacher randomly selects 10 out of her 30 students and finds that the mean height of those 10 students is 5′2". Is this a samp
    14·1 answer
  • Please help me with my HW
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!