All Answers 358108 Answers

... that's so easy, just go plug it into a calculator and you'll get the answer. 

The answer is not 69, that's not funny. Also don't impersonate GA, THE REAL ANSWER IS 4. 

For the record, Post # 5, signed by GA, is not by the real GA.

sin(theta) = cos(5) - sin(25) = sin(85) - sin(25).

Now use the trig identity for  sin(A) - sin(B).

For the record, Post # 14, signed by GA, is not by the real GA.

To solve this problem, we can use the ratio of the tangent of the sun's elevation angle
\(tan x = \frac{35}{50}\)

To find x, we need to apply the inverse tangent function to both sides of the equation

\(x =arctan\frac{35}{50}\)

Using the calculator, we get:

\(x ={34.99}^{o}\)

Rounding up to the nearest power, we get:

\(x ={35}^{o}\)

Answer: The angle of elevation of the sun is approximately 35 degrees.

Let's first find the resistance of the wire using Ohm's formula:

R = V/I = 60/0.2 = 300 ohm

Now, if we cut the wire in half, its length will be halved, which means that the resistance will also be halved:

R/2 = 300/2 = 150 ohm

Now we can find the current that will flow through half the wire at 400 volts, again using Ohm's formula:

I = V/R = 400/150 = 2.67 amps

In milliamps, this would be:

I = 2.67 * 1000 = 2670 milliamps

Answer: At 400 volts, 2670 milliamps of current will flow through half of the wire.

You need to isolate a on one side of the equation and b on the other side. 

3a+4b=5a−6b+24−3a
Combine 5a and -3a to get 2a.
3a+4b=2a−6b+24
Subtract 2a from both sides of the equation.
3a+4b−2a=−6b+24
Combine 3a and −2a to get a.
a+4b=−6b+24
Subtract 4b from both sides of the equation.
a=−6b+24−4b
Combine -6b and -4b to get -10b.
a=−10b+24

Therefore, the solution is a = 24 – 10b

Let's first find the resistance of the wire using Ohm's formula:
R = V/I = 60/0.2 = 300 ohm
Now, if we cut the wire in half, its length will be halved, which means that the resistance will also be halved:
R/2 = 300/2 = 150 ohm
Now we can find the current that will flow through half the wire at 400 volts, again using Ohm's formula:
I = V/R = 400/150 = 2.67 amps
In milliamps, this would be:
I = 2.67 * 1000 = 2670 milliamps
Answer: At 400 volts, 2670 milliamps of current will flow through half of the wire.

what does  \boxed{6:5}.] mean

Hi HumanBeing,

I am just confirming Tiggsy's amazing answer (must more efficient than my answer), and that the Guest's comment below his post is incorrect, in fact, n=11/8.

\(\text{Given } \space \space y=ax^2+bx+c \space \space \space \space \space (*) \\ \)

\(\text{If the parabola passes through (1,1) and (4,-7)} \\ \text{Then, they must satisfy (*), thus substituting into the equation yields: } \\ \space\space\space\space\space\space\space\space\space1=a+b+c \space\space\space\space\space\space\space\space\space\space (1) \\ \space\space\space\space-7=16a+4b+c \space\space\space\space(2) \\\)

\( \text{Now, we have a system of two equations but with three unknowns} \\ \text{So, we have to use the last bit of information} \\ \text{Basically, if the greater root is } \space\space\space \sqrt{n} +2 = 2+\sqrt{n} \\ \text{Then, we know that irrational roots come in pairs, I.e. recall the following theorem: } \\ \text{if } \space\space a+\sqrt{b} \space\space\space \text{is a root, then} \space\space\space a-\sqrt{b} \space\space\space \text{must be a root of the polynomial}.\)

\(\text{Hence, } \space\space 2-\sqrt{n} \space\space \text{is a root} \\ \text{Sum of the roots: } \space \space 4 \\ \text{Product of the roots: } \space\space 4-n \\ \text{Apply Vieta's formulae: } -\dfrac{b}{a} = 4 \space\space\space (**) \\ \dfrac{c}{a}=4-n \space\space\space\space (***)\\ \)

Using (**), we can eliminate b from (1) and (2) as follows:
\(b=-4a \\ \text{(1) and (2) becomes: } \space \\ \iff 1=a-4a+c \implies 1=-3a+c\\ \iff -7=16a-16a+c \implies c=-7 \\ \text{Thus, } a=-\dfrac{8}{3}\)

Using (***): 

\(\dfrac{c}{a}=4-n \\ \iff \dfrac{-7}{\dfrac{-8}{3}}=4-n \\ \iff \dfrac{21}{8}=4-n \\ \iff n=1.375= \dfrac{11}{8} \\ \text{Which is the answer}.\)

And once again, an Ao Ps hw problem.

Are you too lazy to find an online resource or something? We won't spoonfeed you everything.

I did. LOL.

I'm sorry, but no one asked you if anyone asked, so you can just leave. 

Btw I was talking to the second guest

Bart had $635 and Ernie had $403. After they shared the cost of a present equally. Ernie had 3/5 the amounted money Bart had. How much did each of them pay for the present?

The person asking the question asked. 

Heyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy hwwwwwwwwwwwwwwwwwwww cheetah! DO IT YOURSELF!!!! TANQ~!

I've witnessed something like this too. I called out an a o p s question that I remembered from when I was in class, and other users agreed with my sentiments and pointed out that the only answer was ChatGPT generated. In the end, all of our comments were deleted except for the GPT answer. I think there is a serious problem with quality control/academic integrity on this website. I'd say that around 30% of problems here are from an online course such as a o ps, but there is little we can do except for refusing to answer them.

This censor is just st*pid, like we can't mention c*hill or br*inly without being reported? It's like the a o p s censor, made for d*mb*ss reasons to protect no one.

And it isn't fair. A lot of posts about cheating have no "real" evidence, but this question is from Beast Acadamy 100%, you can tell because the name Grogg is one of the little monsters. I saw questions like this when I was doing BA problems