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
Assoli18 [71]
3 years ago
5

HELP ME PLEASE I NEED ANSWERS REALLY FAST I'LL GIVE BRAINLIEST TO THE CORRECT ANSWER SO PLEASE HELP ME!!!!

Mathematics
1 answer:
GuDViN [60]3 years ago
3 0

Answer:

B

Step-by-step explanation:

These lines dont cross CD and are not parrellel which is what skew lines are.

You might be interested in
Find the area of the region enclosed by x(t)=t^2-2t, y(t)=sqrt(t), and the y-axis. ...?
zaharov [31]
First find the the value of t where the curve intersects the Y-axis. This is when x = 0. 

x = t^2 - 2t = 0 = t(t - 2) 


So t= 0 and t = 2 

dA = (0 - x)*dy .... Since the curve has negative x in this region 

y = SQRT(t) and dy = [(1/2)/SQRT(t)]dt 

dA = [2t - t^2][(1/2)/SQRT(t)]dt 

dA = [t^(1/2) - (1/2)t^(3/2)]dt 


Integrate to get: A = (2/3)t^(3/2) - (1/5)t^(5/2) 


Now evaluate from t= 0 to t = 2. 


Area = [(2/3)2^(3/2) - (1/5)2^(5/2)] - [0] 
<span>
Area = SQRT(2)[4/3 - 4/5] 
</span><span>
Area = SQRT(2)[8/15) = 0.754


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
3 0
3 years ago
Which location has the lowest altitude?
Galina-37 [17]
85 meters below sea level death valley all so know as b
8 0
3 years ago
it costs $2.48 for a 16 ounce jar of peanut butter and $5.44 for a 40 ounce jar. which jar is the better buy
gogolik [260]
Second one, please mark branliest
7 0
3 years ago
Let us suppose that one of the fireworks is launched from the top of the building with an initial upward velocity of 150 ft/s an
ivanzaharov [21]

Answer:

9.6 seconds

Step-by-step explanation:

The equation for projectile motion in feet is h(t)=-16t^2+v_0t+h_0 where v_0 is the initial velocity in ft/s and h_0 is the initial height in feet.

We are given that the initial upward velocity is v_0=150ft/sec and the initial height is h_0=35ft. Thus, plugging these values into our equation, we get h(t)=-16t^2+150t+35 as our equation for the situation.

The firework will land, assuming it doesn't explode, when h(t)=0, thus:

h(t)=-16t^2+150t+35\\\\0=-16t^2+150t+35\\\\t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\t=\frac{-150\pm\sqrt{150^2-4(-16)(35)}}{2(-16)}\\ \\t=\frac{-150\pm\sqrt{22500+2240}}{-32}\\\\t=\frac{-150\pm\sqrt{24740}}{-32}\\\\t=\frac{-150\pm2\sqrt{6185}}{-32}\\ \\t=\frac{75\pm\sqrt{6185}}{16}\\\\t_1\approx-0.2\\\\t_2\approx9.6

Since time can't be negative, then the firework will land after 9.6 seconds, assuming it doesn't explode at the time of landing.

3 0
3 years ago
Reserved tickets for the football game cost $20 each and general admission tickets cost $12. The total ticket sales brought in $
Leya [2.2K]

so  what you would do is 20 x 12 and see what you get and the subtract it from 900 and then see what number would take you to 900

6 0
3 years ago
Other questions:
  • How to solve 12x=-24
    5·2 answers
  • What is the effect of decreasing the alpha level (for example, from a = .05 to a = .01)? it decreases the probability of a type
    9·1 answer
  • Why must fire extinguishers be routinely maintained?
    5·2 answers
  • The Milky Way galaxy has atleast 200 million stars right half of this number of stars in standard form
    12·2 answers
  • Another 100 XDDDDD take em
    11·2 answers
  • What is this expression?
    9·1 answer
  • You and your friend each deposit $50 in separate savings accounts. Your account earns 2% simple annual interest Your friend's ac
    14·1 answer
  • Pls help the question below
    15·1 answer
  • If g(x)=x^2-5, determine g(h+1)
    6·1 answer
  • Solve the right triangle. Round decimals to the nearest tenth
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!